Tutorial: Nonmonotonic Logic (Day 1) Christian Straßer Institute for Philosophy II, Ruhr-University Bochum Center for Logic and Philosophy of Science, Ghent University http://homepage.ruhr-uni-bochum.de/defeasible-reasoning/index.html
September 2, 2015
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Outline 1
2 3
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Defeasible Reasoning – Some Basic Concepts Some examples to warm up Some conceptual distinctions Nonmonotonic Logic in Context The Dynamics of Defeasible Reasoning Default Logic (in the tradition of Reiter) Warming up Defaults and Default Theories Inferring Alternatives and more examples Meta-Properties Introducing Priorities Summing up Autoepistemic Logic Bibliography Bibliography
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Overview: the whole tutorial
Day 1 1
2
motivation and basic concepts default logic (Reiter, Poole, Horty) Reiter and variants many examples priorities (Horty)
3
autoepistemic logic (Moore, Konolige, Marek) basics
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Overview: the whole tutorial Day 2 Day 1 1
2
1
motivation and basic concepts default logic (Reiter, Poole, Horty)
default assumptions, adaptive logics reasoning with maximal consistent subsets bridging to preferential semantics meta-theory: interesting properties for non-monotonic logic
Reiter and variants many examples priorities (Horty) 3
autoepistemic logic (Moore, Konolige, Marek) basics
Christian Straßer (RUB, UGENT)
Plausible Reasoning (Makinson, Rescher/Manor, Batens, Geffner/Pearl)
2
Preferential Semantics (Kraus/Lehmann/Magidor, Shoham, Batens) properties, limitations, variants enhancements (rational closure)
Tutorial: Nonmonotonic Logic (Day 1)
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Overview: the whole tutorial Day 2 Day 1 1
2
1
motivation and basic concepts default logic (Reiter, Poole, Horty)
default assumptions, adaptive logics reasoning with maximal consistent subsets bridging to preferential semantics meta-theory: interesting properties for non-monotonic logic
Reiter and variants many examples priorities (Horty) 3
autoepistemic logic (Moore, Konolige, Marek)
Plausible Reasoning (Makinson, Rescher/Manor, Batens, Geffner/Pearl)
2
basics
Preferential Semantics (Kraus/Lehmann/Magidor, Shoham, Batens) properties, limitations, variants enhancements (rational closure)
Yes, . . . I know: it’s not realistic! Christian Straßer (RUB, UGENT)
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Golden Rule
interrupt me — ask — complain — comment
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Slides
can be found at http://homepage.ruhr-uni-bochum.de/defeasible-reasoning/ Courses/Natal2015-NonMonLog/natal15-nonmonlog.html (already outdated . . . )
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Topic 1
2 3
4 5
Defeasible Reasoning – Some Basic Concepts Some examples to warm up Some conceptual distinctions Nonmonotonic Logic in Context The Dynamics of Defeasible Reasoning Default Logic (in the tradition of Reiter) Warming up Defaults and Default Theories Inferring Alternatives and more examples Meta-Properties Introducing Priorities Summing up Autoepistemic Logic Bibliography Bibliography
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What’s defeasible reasoning: some examples
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What’s defeasible reasoning: some examples
abductive inference infer a (good!?) explanation α → β and β thus α Christian Straßer (RUB, UGENT)
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closed world assumption reasoning on the assumption that the given information is complete Christian Straßer (RUB, UGENT)
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Tweety is a bird. Thus, . . . ?
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Tweety is a bird. Thus, . . . ? stereotypical / default reasoning jump to a conclusion on the basis of what is usually/typically/normally/etc. the case
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inductive generalisations
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Domains of defeasible reasoning
everyday reasoning expert reasoning (e.g. medical diagnosis) scientific reasoning
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Commonalities
tentative conclusions → jumping to conclusions retraction possible if problems arise
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Commonalities
tentative conclusions → jumping to conclusions retraction possible if problems arise
Two tiers of defeasible reasoning 1
illative tier (support, concluding)
2
dialectic tier (retraction)
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Ampliative vs. Corrective approaches
Corrective approaches In contrast to ampliative reasoning, each inference is in accordance with CL (or another deductive standard) and hence deductive. However, given an inconsistent theory, not all deductive inferences will be accepted. illative tier: strictly deductive (e.g., classical logic) dialectic tier: conflicting deductive inferences
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Ampliative vs. Corrective approaches
Corrective approaches In contrast to ampliative reasoning, each inference is in accordance with CL (or another deductive standard) and hence deductive. However, given an inconsistent theory, not all deductive inferences will be accepted. illative tier: strictly deductive (e.g., classical logic) dialectic tier: conflicting deductive inferences
Examples nonmonotonic paraconsistent logics (Rescher/Manor, inconsistency-adaptive logics) deductive argumentation-based approaches
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Example: Rescher and Manor’s Free Consequences suppose we reason classically on the basis of a complex body of premises Γ (e.g., a mathematical or scientific theory, or code of law)
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Example: Rescher and Manor’s Free Consequences suppose we reason classically on the basis of a complex body of premises Γ (e.g., a mathematical or scientific theory, or code of law) we do not know whether Γ is consistent
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Example: Rescher and Manor’s Free Consequences suppose we reason classically on the basis of a complex body of premises Γ (e.g., a mathematical or scientific theory, or code of law) we do not know whether Γ is consistent careful rationale for drawing inferences
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Example: Rescher and Manor’s Free Consequences suppose we reason classically on the basis of a complex body of premises Γ (e.g., a mathematical or scientific theory, or code of law) we do not know whether Γ is consistent careful rationale for drawing inferences call a formula ϕ in Γ free if it does not belong to a minimally inconsistent set
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Example: Rescher and Manor’s Free Consequences suppose we reason classically on the basis of a complex body of premises Γ (e.g., a mathematical or scientific theory, or code of law) we do not know whether Γ is consistent careful rationale for drawing inferences call a formula ϕ in Γ free if it does not belong to a minimally inconsistent set rationale: an inference in CL is retracted as soon as we find out that it relies on premises that are not free
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Example: Rescher and Manor’s Free Consequences suppose we reason classically on the basis of a complex body of premises Γ (e.g., a mathematical or scientific theory, or code of law) we do not know whether Γ is consistent careful rationale for drawing inferences call a formula ϕ in Γ free if it does not belong to a minimally inconsistent set rationale: an inference in CL is retracted as soon as we find out that it relies on premises that are not free
Task: Free consequences What are the free consequences of Γ = {p ∧ q, ¬p, r ∧ s}?
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Ampliative approaches While the truth of the premises does not warrant the truth of the conclusion as in deductive reasoning, the conclusion nevertheless holds in most/typical/etc. cases in which the premises hold.
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Ampliative approaches While the truth of the premises does not warrant the truth of the conclusion as in deductive reasoning, the conclusion nevertheless holds in most/typical/etc. cases in which the premises hold. illative tier: beyond truth-preservation typically: fixed minimal deductive (non-defeasible) standard of reasoning / core rules (e.g., classical logic) jump to more conclusions given additional warrants that allow for defeasible conclusions
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Ampliative approaches While the truth of the premises does not warrant the truth of the conclusion as in deductive reasoning, the conclusion nevertheless holds in most/typical/etc. cases in which the premises hold. illative tier: beyond truth-preservation typically: fixed minimal deductive (non-defeasible) standard of reasoning / core rules (e.g., classical logic) jump to more conclusions given additional warrants that allow for defeasible conclusions
dialectic tier: e.g., exceptional circumstances conflicting defeasible and deductive inferences/arguments
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Ampliative approaches While the truth of the premises does not warrant the truth of the conclusion as in deductive reasoning, the conclusion nevertheless holds in most/typical/etc. cases in which the premises hold. illative tier: beyond truth-preservation typically: fixed minimal deductive (non-defeasible) standard of reasoning / core rules (e.g., classical logic) jump to more conclusions given additional warrants that allow for defeasible conclusions
dialectic tier: e.g., exceptional circumstances conflicting defeasible and deductive inferences/arguments
Examples inheritance networks default logic abductive logics inductive generalisation Christian Straßer (RUB, UGENT)
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Classical Logic only illative tier What’s the inference ticket in classical logic (in short, CL)? β follows from α iff in all classical interpretations in which α is true, also β is true entailment: ’truth-preservation’, ’deduction’ e.g. ∀x(P(x) → Q(x)) P(a) Q(a)
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Classical Logic only illative tier What’s the inference ticket in classical logic (in short, CL)? β follows from α iff in all classical interpretations in which α is true, also β is true entailment: ’truth-preservation’, ’deduction’ e.g. ∀x(P(x) → Q(x)) P(a) Q(a)
Hence . . . CL seems not apt to characterize reasoning that is dynamic in the sense that reasoners are prepared to retract inferences. Christian Straßer (RUB, UGENT)
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Summing up
corrective approach ampliative approach
Christian Straßer (RUB, UGENT)
illative tier deductive inferences ded. + non-ded. inf.
dialectic tier conflicting ded. inferencces confl. non-ded. inferences
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Summing up
corrective approach ampliative approach
illative tier deductive inferences ded. + non-ded. inf.
dialectic tier conflicting ded. inferencces confl. non-ded. inferences
In practice: distinction is not so clear-cut
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Pessimism in the 60ies
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Toulmin Scheme (Toulmin (1958))
Premises
Backing
Christian Straßer (RUB, UGENT)
Conclusion
Warrant
Defeat
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A Toulmin Argument
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Nonmonotonic Logic to the Rescue
Figure: Artificial Intelligence, Volume 13, Issues 1–2, Pages 1-174,(April 1980), Special Issue on Non-Monotonic Logic
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Nonmonotonic Logic to the Rescue
Figure: Artificial Intelligence, Volume 13, Issues 1–2, Pages 1-174,(April 1980), Special Issue on Non-Monotonic Logic
Aim capture defeasible reasoning in a mathematically precise way reproduce the success of CL in the domain of mathematical reasoning in the less sterile/idealized domain of defeasible reasoning where incompleteness and uncertainty play a central role Christian Straßer (RUB, UGENT)
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Nowadays
cooperations between formal and informal logicians e.g., Douglas Walton and formal argumentation (Gordon et al. (2007))
shift of normative standards in cognitive science, e.g. Stenning and Van Lambalgen (2008) Pfeifer (2014)
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Topic 1
2 3
4 5
Defeasible Reasoning – Some Basic Concepts Some examples to warm up Some conceptual distinctions Nonmonotonic Logic in Context The Dynamics of Defeasible Reasoning Default Logic (in the tradition of Reiter) Warming up Defaults and Default Theories Inferring Alternatives and more examples Meta-Properties Introducing Priorities Summing up Autoepistemic Logic Bibliography Bibliography
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Nonmonotonicity Nonmonotonic Logics Monotony: If Γ |∼ φ then Γ ∪ Γ0 |∼ φ.
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Nonmonotonicity Nonmonotonic Logics Monotony: If Γ |∼ φ then Γ ∪ Γ0 |∼ φ. If we define a consequence function Cn(•) by Cn(Γ) = {φ | Γ |∼ φ} we can equivalently express Monotony by: If φ ∈ Cn(Γ) then φ ∈ Cn(Γ ∪ Γ0 ).
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Nonmonotonicity Nonmonotonic Logics Monotony: If Γ |∼ φ then Γ ∪ Γ0 |∼ φ. If we define a consequence function Cn(•) by Cn(Γ) = {φ | Γ |∼ φ} we can equivalently express Monotony by: If φ ∈ Cn(Γ) then φ ∈ Cn(Γ ∪ Γ0 ).
External and Internal dynamics (Pollock / Batens) External Dynamics retract conclusions under the influence of new information
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Nonmonotonicity Nonmonotonic Logics Monotony: If Γ |∼ φ then Γ ∪ Γ0 |∼ φ. If we define a consequence function Cn(•) by Cn(Γ) = {φ | Γ |∼ φ} we can equivalently express Monotony by: If φ ∈ Cn(Γ) then φ ∈ Cn(Γ ∪ Γ0 ).
External and Internal dynamics (Pollock / Batens) External Dynamics retract conclusions under the influence of new information
Internal Dynamics retract conclusions under the influence of a progressive analysis of the given information Christian Straßer (RUB, UGENT)
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What to replace Monotonicity with? Cautious Monotony If Γ |∼ ϕ and Γ |∼ ψ, then Γ, ϕ |∼ ψ.
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What to replace Monotonicity with? Cautious Monotony If Γ |∼ ϕ and Γ |∼ ψ, then Γ, ϕ |∼ ψ. Cautious Monotony is the converse of Cut:
(Cautious) Cut If Γ |∼ ϕ and Γ, ϕ |∼ ψ then Γ |∼ ψ.
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What to replace Monotonicity with? Cautious Monotony If Γ |∼ ϕ and Γ |∼ ψ, then Γ, ϕ |∼ ψ. Cautious Monotony is the converse of Cut:
(Cautious) Cut If Γ |∼ ϕ and Γ, ϕ |∼ ψ then Γ |∼ ψ. Cautious Monotonicity (Cut) states that adding a consequence ϕ back into the premise-set Γ does not lead to any decrease (increase) in inferential power.
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What to replace Monotonicity with? Cautious Monotony If Γ |∼ ϕ and Γ |∼ ψ, then Γ, ϕ |∼ ψ. Cautious Monotony is the converse of Cut:
(Cautious) Cut If Γ |∼ ϕ and Γ, ϕ |∼ ψ then Γ |∼ ψ. Cautious Monotonicity (Cut) states that adding a consequence ϕ back into the premise-set Γ does not lead to any decrease (increase) in inferential power. Both together tell us that inference is a cumulative enterprise: we can keep drawing consequences that can in turn be used as additional premises, without affecting the set of conclusions. Christian Straßer (RUB, UGENT)
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Core property
central place in non-monotonic logic Gabbay Gabbay (1985) Kraus, Lehmann, Magidor ("KLM") Kraus et al. (1990)
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Core property
central place in non-monotonic logic Gabbay Gabbay (1985) Kraus, Lehmann, Magidor ("KLM") Kraus et al. (1990)
also empirically confirmed property of actual reasoning
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Core property
central place in non-monotonic logic Gabbay Gabbay (1985) Kraus, Lehmann, Magidor ("KLM") Kraus et al. (1990)
also empirically confirmed property of actual reasoning however: misses e.g., from Reiter’s Default Logic (see below)
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Be Cautious with Cut
(Cautious) Cut If Γ |∼ ϕ and Γ, ϕ |∼ ψ then Γ |∼ ψ.
. . . to be distinguished from: If Γ0 |∼ ϕ and Γ, ϕ |∼ ψ then Γ, Γ0 |∼ ψ.
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Be Cautious with Cut
(Cautious) Cut If Γ |∼ ϕ and Γ, ϕ |∼ ψ then Γ |∼ ψ.
. . . to be distinguished from: If Γ0 |∼ ϕ and Γ, ϕ |∼ ψ then Γ, Γ0 |∼ ψ. Do you see why the second version of CUT is not suitable for nonmonotonic logic?
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What to replace Monotonicity with?
Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ.
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What to replace Monotonicity with?
Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ. Adding formulas that are consistent with our current beliefs does not lead to a decrease in our set of conclusions.
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Stalnaker’s problematic example for Rational Monotonicity Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ.
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Stalnaker’s problematic example for Rational Monotonicity Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ.
Stalnaker’s counter-example (Stalnaker (1994)) Stalnaker adopts the reading of ’α |∼ β ’ as ’Given our initial set of beliefs Γ, if we learn α then (nonmonotonically) infer β.’ proposed in Makinson and Gärdenfors (1991).
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Stalnaker’s problematic example for Rational Monotonicity Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ.
Stalnaker’s counter-example (Stalnaker (1994)) Stalnaker adopts the reading of ’α |∼ β ’ as ’Given our initial set of beliefs Γ, if we learn α then (nonmonotonically) infer β.’ proposed in Makinson and Gärdenfors (1991). Consider the three composers: Verdi (v), Bizet (b), and Satie (s), and suppose that we initially accept (correctly but defeasibly) that Verdi is Italian (I(v)), while Bizet and Satie are French (F(b), F(s)).
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Stalnaker’s problematic example for Rational Monotonicity Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ.
Stalnaker’s counter-example (Stalnaker (1994)) Stalnaker adopts the reading of ’α |∼ β ’ as ’Given our initial set of beliefs Γ, if we learn α then (nonmonotonically) infer β.’ proposed in Makinson and Gärdenfors (1991). Consider the three composers: Verdi (v), Bizet (b), and Satie (s), and suppose that we initially accept (correctly but defeasibly) that Verdi is Italian (I(v)), while Bizet and Satie are French (F(b), F(s)). Suppose now that we learn that Verdi and Bizet are compatriots (C(v,b)).
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Stalnaker’s problematic example for Rational Monotonicity Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ.
Stalnaker’s counter-example (Stalnaker (1994)) Stalnaker adopts the reading of ’α |∼ β ’ as ’Given our initial set of beliefs Γ, if we learn α then (nonmonotonically) infer β.’ proposed in Makinson and Gärdenfors (1991). Consider the three composers: Verdi (v), Bizet (b), and Satie (s), and suppose that we initially accept (correctly but defeasibly) that Verdi is Italian (I(v)), while Bizet and Satie are French (F(b), F(s)). Suppose now that we learn that Verdi and Bizet are compatriots (C(v,b)). then: C (v , b) |∼ F (s)
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Stalnaker’s problematic example for Rational Monotonicity Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ.
Stalnaker’s counter-example (Stalnaker (1994)) Stalnaker adopts the reading of ’α |∼ β ’ as ’Given our initial set of beliefs Γ, if we learn α then (nonmonotonically) infer β.’ proposed in Makinson and Gärdenfors (1991). Consider the three composers: Verdi (v), Bizet (b), and Satie (s), and suppose that we initially accept (correctly but defeasibly) that Verdi is Italian (I(v)), while Bizet and Satie are French (F(b), F(s)). Suppose now that we learn that Verdi and Bizet are compatriots (C(v,b)). then: C (v , b) |∼ F (s) Now consider: C(v,s), then C (v , b) 6|∼ ¬C (v , s). Christian Straßer (RUB, UGENT)
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Stalnaker’s problematic example for Rational Monotonicity Rational Monotony If it’s not the case that Γ |∼ ¬ϕ, and moreover Γ |∼ ψ, then Γ, ϕ |∼ ψ.
Stalnaker’s counter-example (Stalnaker (1994)) Stalnaker adopts the reading of ’α |∼ β ’ as ’Given our initial set of beliefs Γ, if we learn α then (nonmonotonically) infer β.’ proposed in Makinson and Gärdenfors (1991). Consider the three composers: Verdi (v), Bizet (b), and Satie (s), and suppose that we initially accept (correctly but defeasibly) that Verdi is Italian (I(v)), while Bizet and Satie are French (F(b), F(s)). Suppose now that we learn that Verdi and Bizet are compatriots (C(v,b)). then: C (v , b) |∼ F (s) Now consider: C(v,s), then C (v , b) 6|∼ ¬C (v , s). But if we add C(v,s) to our beliefs, then C (v , b), C (v , s) 6|∼ F (s). Christian Straßer (RUB, UGENT)
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Schematic Inference Graphs (e.g., Inheritance Nets, Formal Argumentation, etc.)
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Schematic Inference Graphs (e.g., Inheritance Nets, Formal Argumentation, etc.)
We use the following conventions: →: signify deductive or strict (i.e., non-defeasible) inferences, →: signify defeasible inferences, and
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Schematic Inference Graphs (e.g., Inheritance Nets, Formal Argumentation, etc.)
We use the following conventions: →: signify deductive or strict (i.e., non-defeasible) inferences, →: signify defeasible inferences, and strikethrough (single resp. double) arrows signify that the negation of the pointed formula is (defeasibly resp. strictly) implied.
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Schematic Inference Graphs (e.g., Inheritance Nets, Formal Argumentation, etc.)
We use the following conventions: →: signify deductive or strict (i.e., non-defeasible) inferences, →: signify defeasible inferences, and strikethrough (single resp. double) arrows signify that the negation of the pointed formula is (defeasibly resp. strictly) implied. Arguments: e.g., Penguin → Bird → flies
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Schematic Inference Graphs (e.g., Inheritance Nets, Formal Argumentation, etc.)
We use the following conventions: →: signify deductive or strict (i.e., non-defeasible) inferences, →: signify defeasible inferences, and strikethrough (single resp. double) arrows signify that the negation of the pointed formula is (defeasibly resp. strictly) implied. Arguments: e.g., Penguin → Bird → flies So, we can read the diagram as follows: Penguins are birds (no exceptions); Birds usually fly; and Penguins usually don’t fly. Christian Straßer (RUB, UGENT)
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Resolution: Strict beats defeasible
Two types of conflicts 1
2
conflicts between defeasible conclusions and "hard facts,": A → B vs. A0 → ¬B conflicts between one potential defeasible conclusion and another: A → B vs. A0 → ¬B
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Resolution: Strict beats defeasible
Two types of conflicts 1
2
conflicts between defeasible conclusions and "hard facts,": A → B vs. A0 → ¬B conflicts between one potential defeasible conclusion and another: A → B vs. A0 → ¬B
Resolution Concerning 1: hard facts are prioritized. Concerning 2: this is more complicated . . .
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Specificity / Preemption
According to the Specificity Principle an inference with a more specific antecedent overrides a conflicting defeasible inference with a less specific antecedent.
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Specificity / Preemption
According to the Specificity Principle an inference with a more specific antecedent overrides a conflicting defeasible inference with a less specific antecedent. lot of work in inheritance nets is devoted to this problem (see Horty (1994))
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Topic 1
2 3
4 5
Defeasible Reasoning – Some Basic Concepts Some examples to warm up Some conceptual distinctions Nonmonotonic Logic in Context The Dynamics of Defeasible Reasoning Default Logic (in the tradition of Reiter) Warming up Defaults and Default Theories Inferring Alternatives and more examples Meta-Properties Introducing Priorities Summing up Autoepistemic Logic Bibliography Bibliography
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Some References to Classical Articles
A logic for default reasoning. Artificial Intelligence, 1–2(13). Reiter (1980) A logical framework for default reasoning. Artificial intelligence, 36(1), 27–47. Poole (1988) The effect of knowledge on belief: conditioning, specificity and the lottery paradox in default reasoning. Artificial Intelligence, 49(1-3), 281–307. Poole (1991) Considerations on default logic: an alternative approach. Computational intelligence, 4(1), 1–16. Łukaszewicz (1988)
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Short Reminder: 1st order logic
Logical symbols quantifiers ∀, ∃ logical connectives ∧, ∨, ⊃, ¬ brackets variables
non-logical symbols predicate / relation symbols with specific arity function symbols with specific arity constants (0-ary functions)
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Short Reminder: 1st order logic, special terminology terms: variables, f (t1 , . . . , tn ) where ti are terms atomic formula: P(t1 , . . . , tn ) formulas: h∀, ∃, ∧, ∨, ⊃, ¬i closure of atomic formulas free / bound variables sentence: formula without free variables instance of a formula ϕ: substitution of some free variables for terms ground term: term without variables ground instance: instance that is a sentence (obtained by substituting all free variables by ground terms)
Example bird(Tweety) ⊃ flies(Tweety) is a ground instance of bird(x) ⊃ flies(x) Christian Straßer (RUB, UGENT)
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What’s a default conditional prerequisite α(x) : β1 (x), . . . , βn (b) γ(x)
where x = x1 , . . . , xm , and α(x), β1 (x), . . . , βn (x), γ(x) are formulas whose free variables are among x1 , . . . , xm .
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What’s a default conditional prerequisite
justification
α(x) : β1 (x), . . . , βn (b) γ(x)
where x = x1 , . . . , xm , and α(x), β1 (x), . . . , βn (x), γ(x) are formulas whose free variables are among x1 , . . . , xm .
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What’s a default conditional prerequisite
justification
α(x) : β1 (x), . . . , βn (b) γ(x) conclusion where x = x1 , . . . , xm , and α(x), β1 (x), . . . , βn (x), γ(x) are formulas whose free variables are among x1 , . . . , xm .
Christian Straßer (RUB, UGENT)
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What’s a default conditional prerequisite
justification
α(x) : β1 (x), . . . , βn (b) γ(x) conclusion where x = x1 , . . . , xm , and α(x), β1 (x), . . . , βn (x), γ(x) are formulas whose free variables are among x1 , . . . , xm .
Application of a default The default is applied in order to derive the c-ground instance of γ in case trigger: α(c) belongs to our set of —depending on the perspective we have— beliefs/(defeasible) knowledge/plausible assumptions/etc. (henceforth I will speak only about beliefs) justification: the set of our beliefs is consistent with each βi (c) Christian Straßer (RUB, UGENT)
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Default Theory
h∆, Φi set of defaults
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Default Theory
set of ’facts’ h∆, Φi set of defaults
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Default Theory
set of ’facts’ h∆, Φi set of defaults
simple example
bird(x) : flies(x) flies(x) Φ = {bird(Tweety), cat(Sylvester)} ∆=
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Types of defaults Normal defaults α(x) : γ(x) γ(x) rather natural representation
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Types of defaults Normal defaults α(x) : γ(x) γ(x) rather natural representation
Semi-Normal defaults α(x) : β(x) γ(x) where β(x) ` γ(x). E.g., α(x) : γ(x) ∧ β(x) γ(x) Christian Straßer (RUB, UGENT)
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How to reason with default theories?
Idea Apply iteratively modus ponens to defaults. This way build step-wise an extension (sets of beliefs that are obtained in this way)
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Here’s how it goes: guess the extension Ξ
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether:
Christian Straßer (RUB, UGENT)
α(x) : β(x) γ(x)
Tutorial: Nonmonotonic Logic (Day 1)
∈ ∆ and
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c)
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)}
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no:
Christian Straßer (RUB, UGENT)
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no: try another triggered default in ∆ (goto (†))
Christian Straßer (RUB, UGENT)
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no: try another triggered default in ∆ (goto (†)) if there isn’t: terminate.
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no: try another triggered default in ∆ (goto (†)) if there isn’t: terminate. if Ξ = Cn(Ξ? ): extension found.
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Here’s how it goes: guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no: try another triggered default in ∆ (goto (†)) if there isn’t: terminate. if Ξ = Cn(Ξ? ): extension found.
Problem (?) We have to guess and use our guess when adding new defaults. Christian Straßer (RUB, UGENT)
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x)
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x) and Φ = {bird(Tweety), cat(Sylvester)}.
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x) and Φ = {bird(Tweety), cat(Sylvester)}. we have only the two constants Tweety and Sylvester in the language
Christian Straßer (RUB, UGENT)
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x) and Φ = {bird(Tweety), cat(Sylvester)}. we have only the two constants Tweety and Sylvester in the language
Building up the extensions: guess: Ξ = Cn({flies(Tweety)} ∪ Φ})
Christian Straßer (RUB, UGENT)
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x) and Φ = {bird(Tweety), cat(Sylvester)}. we have only the two constants Tweety and Sylvester in the language
Building up the extensions: guess: Ξ = Cn({flies(Tweety)} ∪ Φ}) our initial knowledge is Φ
Christian Straßer (RUB, UGENT)
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x) and Φ = {bird(Tweety), cat(Sylvester)}. we have only the two constants Tweety and Sylvester in the language
Building up the extensions: guess: Ξ = Cn({flies(Tweety)} ∪ Φ}) our initial knowledge is Φ note that the Sylvester-instance of our default is not applicable to Φ since Φ 0 bird(Sylvester)
Christian Straßer (RUB, UGENT)
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x) and Φ = {bird(Tweety), cat(Sylvester)}. we have only the two constants Tweety and Sylvester in the language
Building up the extensions: guess: Ξ = Cn({flies(Tweety)} ∪ Φ}) our initial knowledge is Φ note that the Sylvester-instance of our default is not applicable to Φ since Φ 0 bird(Sylvester) however, we have bird(Tweety) and fly(Tweety) is consistent with Ξ.
Christian Straßer (RUB, UGENT)
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x) and Φ = {bird(Tweety), cat(Sylvester)}. we have only the two constants Tweety and Sylvester in the language
Building up the extensions: guess: Ξ = Cn({flies(Tweety)} ∪ Φ}) our initial knowledge is Φ note that the Sylvester-instance of our default is not applicable to Φ since Φ 0 bird(Sylvester) however, we have bird(Tweety) and fly(Tweety) is consistent with Ξ. fixed point reached
Christian Straßer (RUB, UGENT)
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Example: Tweety Let T = h∆, Φi where bird(x) : flies(x) ∆= flies(x) and Φ = {bird(Tweety), cat(Sylvester)}. we have only the two constants Tweety and Sylvester in the language
Building up the extensions: guess: Ξ = Cn({flies(Tweety)} ∪ Φ}) our initial knowledge is Φ note that the Sylvester-instance of our default is not applicable to Φ since Φ 0 bird(Sylvester) however, we have bird(Tweety) and fly(Tweety) is consistent with Ξ. fixed point reached the only extension is Ξ. Christian Straßer (RUB, UGENT)
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Question
Are extensions always unique?
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The Nixon Diamond
Let T = h∆, Φi where ∆= n
quaker(x) : pacifist(x) republican(x) : ¬pacifist(x) , pacifist(x) ¬pacifist(x)
o
Φ = {Quaker(Nixon), republican(Nixon)}.
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The Nixon Diamond
Let T = h∆, Φi where ∆= n
quaker(x) : pacifist(x) republican(x) : ¬pacifist(x) , pacifist(x) ¬pacifist(x)
o
Φ = {Quaker(Nixon), republican(Nixon)}.
There are two extensions:
Christian Straßer (RUB, UGENT)
1
one that contains pacifist(Nixon),
2
and one that contains ¬pacifist(Nixon).
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Another example Let h∆, Φi be a default theory where α(x) : β(x) γ(x) : ¬β(x) , ∆= γ(x) ¬β(x) Φ = {α(c)}
Question Is there an extension?
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Another example Let h∆, Φi be a default theory where α(x) : β(x) γ(x) : ¬β(x) , ∆= γ(x) ¬β(x) Φ = {α(c)}
Question Is there an extension?
Guess: Cn({α(c), γ(c)})
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Another example Let h∆, Φi be a default theory where α(x) : β(x) γ(x) : ¬β(x) , ∆= γ(x) ¬β(x) Φ = {α(c)}
Question Is there an extension?
Guess: Cn({α(c), γ(c)}) first run: Φ? = Φ
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Another example Let h∆, Φi be a default theory where α(x) : β(x) γ(x) : ¬β(x) , ∆= γ(x) ¬β(x) Φ = {α(c)}
Question Is there an extension?
Guess: Cn({α(c), γ(c)}) first run: Φ? = Φ α(x) : β(x) γ(x)
is triggered and justified: apply
Christian Straßer (RUB, UGENT)
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Another example Let h∆, Φi be a default theory where α(x) : β(x) γ(x) : ¬β(x) , ∆= γ(x) ¬β(x) Φ = {α(c)}
Question Is there an extension?
Guess: Cn({α(c), γ(c)}) first run: Φ? = Φ α(x) : β(x) γ(x)
is triggered and justified: apply Φ? = Φ ∪ {γ(c)} Christian Straßer (RUB, UGENT)
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Another example Let h∆, Φi be a default theory where α(x) : β(x) γ(x) : ¬β(x) , ∆= γ(x) ¬β(x) Φ = {α(c)}
Question Is there an extension?
Guess: Cn({α(c), γ(c)}) second run:
first run:
γ(x) : ¬β(x) ¬β(x)
Φ? = Φ α(x) : β(x) γ(x)
is triggered and justified: apply
is triggered and
justified
Φ? = Φ ∪ {γ(c)} Christian Straßer (RUB, UGENT)
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Another example Let h∆, Φi be a default theory where α(x) : β(x) γ(x) : ¬β(x) , ∆= γ(x) ¬β(x) Φ = {α(c)}
Question Is there an extension?
Guess: Cn({α(c), γ(c)}) second run:
first run:
γ(x) : ¬β(x) ¬β(x)
Φ? = Φ α(x) : β(x) γ(x)
is triggered and justified: apply
is triggered and
justified
Φ? = Φ ∪ {γ(c), ¬β(c)}
Φ? = Φ ∪ {γ(c)} Christian Straßer (RUB, UGENT)
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Another example Let h∆, Φi be a default theory where α(x) : β(x) γ(x) : ¬β(x) , ∆= γ(x) ¬β(x) Φ = {α(c)}
Question Is there an extension?
Guess: Cn({α(c), γ(c)}) second run:
first run:
γ(x) : ¬β(x) ¬β(x)
Φ? = Φ α(x) : β(x) γ(x)
is triggered and justified: apply Φ? = Φ ∪ {γ(c)} Christian Straßer (RUB, UGENT)
is triggered and
justified
Φ? = Φ ∪ {γ(c), ¬β(c)} our guess is wrong (similar problems with other guesses)
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How to define the consequences of a default theory?
Two approaches:
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How to define the consequences of a default theory?
Two approaches:
Skeptical approach h∆, Φi `skp A iff A ∈
Christian Straßer (RUB, UGENT)
T
Extensions(h∆, Φi)
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How to define the consequences of a default theory?
Two approaches:
Skeptical approach h∆, Φi `skp A iff A ∈
T
Extensions(h∆, Φi)
S
Extensions(h∆, Φi)
Credulous approach h∆, Φi `crd A iff A ∈
Christian Straßer (RUB, UGENT)
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How to define the consequences of a default theory?
Two approaches:
Skeptical approach h∆, Φi `skp A iff A ∈
T
Extensions(h∆, Φi)
S
Extensions(h∆, Φi)
Credulous approach h∆, Φi `crd A iff A ∈
Question: When is which approach useful?
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Alternative: Consistency Check in the end guess the extension Ξ
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Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ
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Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether:
Christian Straßer (RUB, UGENT)
α(x) : β(x) γ(x)
Tutorial: Nonmonotonic Logic (Day 1)
∈ ∆ and
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Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c)
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Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
Christian Straßer (RUB, UGENT)
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Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)}
Christian Straßer (RUB, UGENT)
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Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no:
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Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no:
try another triggered default in ∆ (goto (†))
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no:
try another triggered default in ∆ (goto (†)) if there isn’t: terminate
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
47 / 104
Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no:
try another triggered default in ∆ (goto (†)) if there isn’t: terminate
if Ξ? is consistent with all the justifications of the applied defaults and if Ξ = Cn(Ξ? ) we found an extension, otherwise try again from the start.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
47 / 104
Alternative: Consistency Check in the end guess the extension Ξ init beliefs: Ξ? = Φ (†) take an c-ground instance of an default check whether: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
trigger?: Ξ? ` α(c) conflicted?: each βi (c) (1 ≤ i ≤ n) is consistent with Ξ (!!)
if yes: update beliefs: Ξ? := Ξ? ∪ {γ(c)} if no:
try another triggered default in ∆ (goto (†)) if there isn’t: terminate
if Ξ? is consistent with all the justifications of the applied defaults and if Ξ = Cn(Ξ? ) we found an extension, otherwise try again from the start. no initial guess guessing on-the-fly when we choose which triggered defaults to apply only in the end we see whether we guessed luckily Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ (†) take a c-instance of an arbitrary default check:
Christian Straßer (RUB, UGENT)
α(x) : β(x) γ(x)
Tutorial: Nonmonotonic Logic (Day 1)
∈ ∆ and
September 2, 2015
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Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ (†) take a c-instance of an arbitrary default check: 1
α(x) : β(x) γ(x)
∈ ∆ and
Φ? ` α(c) (trigger)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
48 / 104
Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ (†) take a c-instance of an arbitrary default check: 1 2
α(x) : β(x) γ(x)
∈ ∆ and
Φ? ` α(c) (trigger) each βi (c) is consistent with Φ? (justification 1)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
48 / 104
Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ (†) take a c-instance of an arbitrary default check: 1 2 3
α(x) : β(x) γ(x)
∈ ∆ and
Φ? ` α(c) (trigger) each βi (c) is consistent with Φ? (justification 1) each justification of previously applied defaults is consistent with Φ? ∪ {γ(c)} (justification 2)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
48 / 104
Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ (†) take a c-instance of an arbitrary default check: 1 2 3
α(x) : β(x) γ(x)
∈ ∆ and
Φ? ` α(c) (trigger) each βi (c) is consistent with Φ? (justification 1) each justification of previously applied defaults is consistent with Φ? ∪ {γ(c)} (justification 2)
if yes: Φ? = Φ? ∪ {γ(c)} and goto (†)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
48 / 104
Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ (†) take a c-instance of an arbitrary default check: 1 2 3
α(x) : β(x) γ(x)
∈ ∆ and
Φ? ` α(c) (trigger) each βi (c) is consistent with Φ? (justification 1) each justification of previously applied defaults is consistent with Φ? ∪ {γ(c)} (justification 2)
if yes: Φ? = Φ? ∪ {γ(c)} and goto (†) if no:
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
48 / 104
Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ (†) take a c-instance of an arbitrary default check: 1 2 3
α(x) : β(x) γ(x)
∈ ∆ and
Φ? ` α(c) (trigger) each βi (c) is consistent with Φ? (justification 1) each justification of previously applied defaults is consistent with Φ? ∪ {γ(c)} (justification 2)
if yes: Φ? = Φ? ∪ {γ(c)} and goto (†) if no: if there is another instance of a default in ∆ that wasn’t tasted, goto (†) and test it
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
48 / 104
Going strictly procedural with Łukaszewicz (1988) Let h∆, Φi be a default theory. init: Φ? = Φ (†) take a c-instance of an arbitrary default check: 1 2 3
α(x) : β(x) γ(x)
∈ ∆ and
Φ? ` α(c) (trigger) each βi (c) is consistent with Φ? (justification 1) each justification of previously applied defaults is consistent with Φ? ∪ {γ(c)} (justification 2)
if yes: Φ? = Φ? ∪ {γ(c)} and goto (†) if no: if there is another instance of a default in ∆ that wasn’t tasted, goto (†) and test it otherwise: if Ξ = Cn(Φ? ), we found an extension.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Some properties of the new procedure No guess needed. real procedural character guarantees existence of an extension hence: yields sometimes different results from Reiter’s account
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Some properties of the new procedure No guess needed. real procedural character guarantees existence of an extension hence: yields sometimes different results from Reiter’s account
Question What happens in the new approach when plugging in the default theory h∆, Φi where α(x) : β(x) ∧ γ(x) γ(x) : ¬β(x) ∆= , γ(x) ¬β(x) Φ = {α(c)}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Recall the three conditions 1 2 3
Φ? ` α(c) (trigger) ?
each βi (c) is consistent with Φ (justification 1) each justification of previously applied defaults is consistent with Φ? ∪ {γ(c)} (justification 2)
∆= n
α(x) : β(x)∧γ(x) γ(x) : ¬β(x) , γ(x) ¬β(x)
Φ = {α(c)}
o
Non-Procedural Fixed-Point Characterisations What about the following definition?
Definition: Extension’ Ξ is an extension’ of a default theory h∆, Φi iff it is a minimal set that satisfies the following conditions:
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Non-Procedural Fixed-Point Characterisations What about the following definition?
Definition: Extension’ Ξ is an extension’ of a default theory h∆, Φi iff it is a minimal set that satisfies the following conditions: 1
Φ⊆Ξ
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
50 / 104
Non-Procedural Fixed-Point Characterisations What about the following definition?
Definition: Extension’ Ξ is an extension’ of a default theory h∆, Φi iff it is a minimal set that satisfies the following conditions: 1
Φ⊆Ξ
2
Cn(Ξ) = Ξ (fixed-point)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
50 / 104
Non-Procedural Fixed-Point Characterisations What about the following definition?
Definition: Extension’ Ξ is an extension’ of a default theory h∆, Φi iff it is a minimal set that satisfies the following conditions: 1
Φ⊆Ξ
2
Cn(Ξ) = Ξ (fixed-point)
3
if
α(c) : β(c) γ(c)
is a c-instance of some default in ∆ and
then γ(c) ∈ Ξ
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
50 / 104
Non-Procedural Fixed-Point Characterisations What about the following definition?
Definition: Extension’ Ξ is an extension’ of a default theory h∆, Φi iff it is a minimal set that satisfies the following conditions: 1
Φ⊆Ξ
2
Cn(Ξ) = Ξ (fixed-point)
3
if
α(c) : β(c) γ(c) 1
is a c-instance of some default in ∆ and
α(c) ∈ Ξ (trigger)
then γ(c) ∈ Ξ
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
50 / 104
Non-Procedural Fixed-Point Characterisations What about the following definition?
Definition: Extension’ Ξ is an extension’ of a default theory h∆, Φi iff it is a minimal set that satisfies the following conditions: 1
Φ⊆Ξ
2
Cn(Ξ) = Ξ (fixed-point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ Ξ (trigger) βi (c) is consistent with Ξ for all 1 ≤ i ≤ n (justification)
then γ(c) ∈ Ξ
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
50 / 104
Non-Procedural Fixed-Point Characterisations What about the following definition?
Definition: Extension’ Ξ is an extension’ of a default theory h∆, Φi iff it is a minimal set that satisfies the following conditions: 1
Φ⊆Ξ
2
Cn(Ξ) = Ξ (fixed-point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ Ξ (trigger) βi (c) is consistent with Ξ for all 1 ≤ i ≤ n (justification)
then γ(c) ∈ Ξ
Question Is this equivalent to the procedural approach? Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Difference: Grounding
n o , ∅}i. Take h >:p p Note that Cn({¬p}) is a minimal set satisfying the previous conditions.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Difference: Grounding
n o , ∅}i. Take h >:p p Note that Cn({¬p}) is a minimal set satisfying the previous conditions. However, the only extension is Cn({p}).
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Difference: Grounding
n o , ∅}i. Take h >:p p Note that Cn({¬p}) is a minimal set satisfying the previous conditions. However, the only extension is Cn({p}). We face the
Problem of grounding We expect that all members of the extension can be generated iteratively by chaining and detaching defaults.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Non-Procedural Fixed-Point Characterisations
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying:
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
52 / 104
Non-Procedural Fixed-Point Characterisations
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
52 / 104
Non-Procedural Fixed-Point Characterisations
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
52 / 104
Non-Procedural Fixed-Point Characterisations
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c)
is a c-instance of some default in ∆ and
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
52 / 104
Non-Procedural Fixed-Point Characterisations
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c) 1
is a c-instance of some default in ∆ and
α(c) ∈ πΦ (Γ) (trigger)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
52 / 104
Non-Procedural Fixed-Point Characterisations
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ πΦ (Γ) (trigger) ¬βi (c) ∈ / Γ for all 1 ≤ i ≤ n then γ(c) ∈ πΦ (Γ) (justification).
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
52 / 104
Non-Procedural Fixed-Point Characterisations
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ πΦ (Γ) (trigger) ¬βi (c) ∈ / Γ for all 1 ≤ i ≤ n then γ(c) ∈ πΦ (Γ) (justification).
Definition: Extension A set of formulas Γ is an extension of h∆, Φi iff π(Γ) = Γ.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying:
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
53 / 104
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ πΦ (Γ) (trigger) ¬βi (c) ∈ / Γ for all 1 ≤ i ≤ n then γ(c) ∈ πΦ (Γ) (justification).
Proof of the existence of π(Γ).
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
53 / 104
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ πΦ (Γ) (trigger) ¬βi (c) ∈ / Γ for all 1 ≤ i ≤ n then γ(c) ∈ πΦ (Γ) (justification).
Proof of the existence of π(Γ). Let S be all sets that satisfy (1)–(3). (Note S 6= ∅ since L ∈ S.)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ πΦ (Γ) (trigger) ¬βi (c) ∈ / Γ for all 1 ≤ i ≤ n then γ(c) ∈ πΦ (Γ) (justification).
Proof of the existence of π(Γ). Let S be T all sets that satisfy (1)–(3). (Note S 6= ∅ since L ∈ S.) Let Γ0 = S. We have to show (1)–(3). 1
trivial
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ πΦ (Γ) (trigger) ¬βi (c) ∈ / Γ for all 1 ≤ i ≤ n then γ(c) ∈ πΦ (Γ) (justification).
Proof of the existence of π(Γ). Let S be T all sets that satisfy (1)–(3). (Note S 6= ∅ since L ∈ S.) Let Γ0 = S. We have to show (1)–(3). 1 2
trivial Suppose A ∈ Cn(Γ0 ). Hence (by monotonicity), Γ00 ` A for all Γ00 ∈ S. T Since Cn(Γ00 ) = Γ00 , A ∈ Γ00 . Thus, A ∈ S.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
53 / 104
Let h∆, Φi be a default theory. Define the operator πΦ such that for any set of formulas Γ, πΦ (Γ) the smallest set satisfying: 1
Φ ⊆ πΦ (Γ)
2
πΦ (Γ) = Cn(πΦ (Γ)) (fixed point)
3
if
α(c) : β(c) γ(c) 1 2
is a c-instance of some default in ∆ and
α(c) ∈ πΦ (Γ) (trigger) ¬βi (c) ∈ / Γ for all 1 ≤ i ≤ n then γ(c) ∈ πΦ (Γ) (justification).
Proof of the existence of π(Γ). Let S be T all sets that satisfy (1)–(3). (Note S 6= ∅ since L ∈ S.) Let Γ0 = S. We have to show (1)–(3). 1 2
3
trivial Suppose A ∈ Cn(Γ0 ). Hence (by monotonicity), Γ00 ` A for all Γ00 ∈ S. T Since Cn(Γ00 ) = Γ00 , A ∈ Γ00 . Thus, A ∈ S. Suppose α(c) ∈ Γ0 and ¬βi (c) ∈ / Γ for all i ≤ n. Hence, α(c) ∈ Γ00 for 00 00 all Γ ∈ S and thus γ(c) ∈ Γ . Thus, γ(c) ∈ Γ0 .
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B. What do you think?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
54 / 104
Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B. What do you think?
Lemma (from this Cut follows immediately for skeptical consequence) If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i).
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
54 / 104
Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B. What do you think?
Lemma (from this Cut follows immediately for skeptical consequence) If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i).
Proof Suppose h∆, Φi ` A and let Ξ ∈ Ext(h∆, Φi).
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B. What do you think?
Lemma (from this Cut follows immediately for skeptical consequence) If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i).
Proof Suppose h∆, Φi ` A and let Ξ ∈ Ext(h∆, Φi). We know that Ξ = πΦ (Ξ) and have to show that Ξ = πΦ∪{A} (Ξ).
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B. What do you think?
Lemma (from this Cut follows immediately for skeptical consequence) If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i).
Proof Suppose h∆, Φi ` A and let Ξ ∈ Ext(h∆, Φi). We know that Ξ = πΦ (Ξ) and have to show that Ξ = πΦ∪{A} (Ξ). Clearly, since A ∈ Ξ, Ξ satisfies (1)–(3) (relative to Φ ∪ {A}).
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
54 / 104
Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B. What do you think?
Lemma (from this Cut follows immediately for skeptical consequence) If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i).
Proof Suppose h∆, Φi ` A and let Ξ ∈ Ext(h∆, Φi). We know that Ξ = πΦ (Ξ) and have to show that Ξ = πΦ∪{A} (Ξ). Clearly, since A ∈ Ξ, Ξ satisfies (1)–(3) (relative to Φ ∪ {A}). Assume πΦ∪{A} (Ξ) ⊂ Ξ.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
54 / 104
Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B. What do you think?
Lemma (from this Cut follows immediately for skeptical consequence) If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i).
Proof Suppose h∆, Φi ` A and let Ξ ∈ Ext(h∆, Φi). We know that Ξ = πΦ (Ξ) and have to show that Ξ = πΦ∪{A} (Ξ). Clearly, since A ∈ Ξ, Ξ satisfies (1)–(3) (relative to Φ ∪ {A}). Assume πΦ∪{A} (Ξ) ⊂ Ξ. But then πΦ∪{A} (Ξ) also satisfies (1)–(3) relative to Φ which contradicts Ξ = πΦ (Ξ). Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Cautious Cut If h∆, Φi ` A and h∆, Φ ∪ {A}i ` B then h∆, Φi ` B. What do you think?
Lemma (from this Cut follows immediately for skeptical consequence) If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i).
Proof Suppose h∆, Φi ` A and let Ξ ∈ Ext(h∆, Φi). We know that Ξ = πΦ (Ξ) and have to show that Ξ = πΦ∪{A} (Ξ). Clearly, since A ∈ Ξ, Ξ satisfies (1)–(3) (relative to Φ ∪ {A}). Assume πΦ∪{A} (Ξ) ⊂ Ξ. But then πΦ∪{A} (Ξ) also satisfies (1)–(3) relative to Φ which contradicts Ξ = πΦ (Ξ). Hence, Ξ = πΦ∪{A} (Ξ). Christian Straßer (RUB, UGENT)
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Cautious Cut for credulous version?
Recall: Lemma If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i).
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Cautious Cut for credulous version?
Recall: Lemma If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i). What do you think, does this help?
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Cautious Cut for credulous version?
Recall: Lemma If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i). What do you think, does this help?
Counter-example Take h∆, Φi where ∆ =
n
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
>:p p∨q:¬p p , ¬p
o .
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Cautious Cut for credulous version?
Recall: Lemma If h∆, Φi ` A, Ext(h∆, Φi) ⊆ Ext(h∆, Φ ∪ {A}i). What do you think, does this help?
Counter-example Take h∆, Φi where ∆ =
n
>:p p∨q:¬p p , ¬p
o .
h∆, Φi `cred p ∨ q. h∆, Φ ∪ {p ∨ q}i `cred ¬p. But, h∆, Φi 0cred ¬p.
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The special status of normal default theories
A normal default theory is a default theory that only consists of normal defaults.
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The special status of normal default theories
A normal default theory is a default theory that only consists of normal defaults. A normal default theory always has an extension both in Reiter’s and in Lukaszewicz’s approach.
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The special status of normal default theories
A normal default theory is a default theory that only consists of normal defaults. A normal default theory always has an extension both in Reiter’s and in Lukaszewicz’s approach. For normal theories the set of Reiter extensions and the set of Lukaszewicz extensions coincides.
Christian Straßer (RUB, UGENT)
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Floating conclusions
Task 1
2
Christian Straßer (RUB, UGENT)
What are the two extensions of this default theory? Is politically−motivated(Nixon) derivable?
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{ Nixon, quaker, republican, dove, hawk, politically motivated }
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{ Nixon, quaker, republican, dove, hawk, politically motivated }
Christian Straßer (RUB, UGENT)
{ Nixon, quaker, republican, dove, ¬hawk, politically motivated }
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{ Nixon, quaker, republican, dove, hawk, politically motivated }
Christian Straßer (RUB, UGENT)
{ Nixon, quaker, republican, dove, ¬hawk, politically motivated }
Tutorial: Nonmonotonic Logic (Day 1)
{ Nixon, quaker, republican, ¬dove, hawk, politically motivated }
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Specificity
Question Is flies(Tweety) derivable?
Christian Straßer (RUB, UGENT)
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Specificity
Question Is flies(Tweety) derivable?
Nope There are two extensions:
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Specificity
Question Is flies(Tweety) derivable?
Nope There are two extensions: 1
one with flies(Tweety)
2
one with ¬flies(Tweety)
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Lukaszewicz’s Fishing example Let T = h∆, Φi where ∆= n
Sunday : I−go−fishing∧¬I−wake−up−late Holidays : I−wake−up−late , I−go−fishing I−wake−up−late
o
Φ = {Sunday, Holidays}.
Christian Straßer (RUB, UGENT)
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Lukaszewicz’s Fishing example Let T = h∆, Φi where ∆= n
Sunday : I−go−fishing∧¬I−wake−up−late Holidays : I−wake−up−late , I−go−fishing I−wake−up−late
o
Φ = {Sunday, Holidays}.
Reiter there is only the extension containing Sunday, Holidays, I−wake−up−late (by first applying the second and then the first default)
Christian Straßer (RUB, UGENT)
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Lukaszewicz’s Fishing example Let T = h∆, Φi where ∆= n
Sunday : I−go−fishing∧¬I−wake−up−late Holidays : I−wake−up−late , I−go−fishing I−wake−up−late
o
Φ = {Sunday, Holidays}.
Reiter there is only the extension containing Sunday, Holidays, I−wake−up−late (by first applying the second and then the first default)
Lukaszewicz we also(!) have the extension that is the result of first applying the first default
Christian Straßer (RUB, UGENT)
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Lukaszewicz’s Fishing example Let T = h∆, Φi where ∆= n
Sunday : I−go−fishing∧¬I−wake−up−late Holidays : I−wake−up−late , I−go−fishing I−wake−up−late
o
Φ = {Sunday, Holidays}.
Reiter there is only the extension containing Sunday, Holidays, I−wake−up−late (by first applying the second and then the first default)
Lukaszewicz we also(!) have the extension that is the result of first applying the first default
Question What do you make of it? Christian Straßer (RUB, UGENT)
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Poole’s Lottery Paradox Let T = h∆, Φi where bird(x) : flies(x) ∧ ¬penguin(x) ∆= , flies(x) ∧ ¬penguin(x) bird(x) : treenest(x) ∧ ¬sandpiper(x) ,... treenest(x) ∧ ¬sandpiper(x) Φ = {bird(Tweety)}
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Poole’s Lottery Paradox Let T = h∆, Φi where bird(x) : flies(x) ∧ ¬penguin(x) ∆= , flies(x) ∧ ¬penguin(x) bird(x) : treenest(x) ∧ ¬sandpiper(x) ,... treenest(x) ∧ ¬sandpiper(x) Φ = {bird(Tweety)}
Problem However, then we conclude ¬penguin(x) ∧ ¬sandpiper(x) ∧ ¬ . . . for all bird-species. But then Tweety does not belong to any species of birds. Typical birds (in an ideal sense) do not exist.
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Reflections about Extensions should there always be extensions?
Christian Straßer (RUB, UGENT)
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Reflections about Extensions should there always be extensions? what do extensions represent?
Christian Straßer (RUB, UGENT)
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Reflections about Extensions should there always be extensions? what do extensions represent? equilibrium states of a rational reasoner?
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Reflections about Extensions should there always be extensions? what do extensions represent? equilibrium states of a rational reasoner? “different, possibly conflicting conclusion sets as rational outcomes based on initial information” (Horty, 2005)
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Reflections about Extensions should there always be extensions? what do extensions represent? equilibrium states of a rational reasoner? “different, possibly conflicting conclusion sets as rational outcomes based on initial information” (Horty, 2005) good reasons approach: if A is in an extension then there are good reasons to suppose A (see Nixon)
Christian Straßer (RUB, UGENT)
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Reflections about Extensions should there always be extensions? what do extensions represent? equilibrium states of a rational reasoner? “different, possibly conflicting conclusion sets as rational outcomes based on initial information” (Horty, 2005) good reasons approach: if A is in an extension then there are good reasons to suppose A (see Nixon)
are some extensions preferable to others?
Christian Straßer (RUB, UGENT)
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Reflections about Extensions should there always be extensions? what do extensions represent? equilibrium states of a rational reasoner? “different, possibly conflicting conclusion sets as rational outcomes based on initial information” (Horty, 2005) good reasons approach: if A is in an extension then there are good reasons to suppose A (see Nixon)
are some extensions preferable to others? should some extensions be filtered out?
Christian Straßer (RUB, UGENT)
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Reflections about Extensions should there always be extensions? what do extensions represent? equilibrium states of a rational reasoner? “different, possibly conflicting conclusion sets as rational outcomes based on initial information” (Horty, 2005) good reasons approach: if A is in an extension then there are good reasons to suppose A (see Nixon)
are some extensions preferable to others? should some extensions be filtered out? how to build extensions (naturally) in order to explicate actual default reasoning?
Christian Straßer (RUB, UGENT)
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Reflections about Extensions should there always be extensions? what do extensions represent? equilibrium states of a rational reasoner? “different, possibly conflicting conclusion sets as rational outcomes based on initial information” (Horty, 2005) good reasons approach: if A is in an extension then there are good reasons to suppose A (see Nixon)
are some extensions preferable to others? should some extensions be filtered out? how to build extensions (naturally) in order to explicate actual default reasoning? should floating conclusions be accepted?
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Problems with Disjunctions
Let T = h∆, Φi where Quaker(x) : dove(x) republican(x) : hawk(x) , , ∆= dove(x) hawk(x)
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Problems with Disjunctions
Let T = h∆, Φi where Quaker(x) : dove(x) republican(x) : hawk(x) , , ∆= dove(x) hawk(x) Φ = {Quaker(Peter) ∨ republican(Peter), Quaker(Anne) ∨ Quaker(George)}.
Christian Straßer (RUB, UGENT)
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Problems with Disjunctions
Let T = h∆, Φi where Quaker(x) : dove(x) republican(x) : hawk(x) , , ∆= dove(x) hawk(x) Φ = {Quaker(Peter) ∨ republican(Peter), Quaker(Anne) ∨ Quaker(George)}.
Problem we don’t get hawk(Peter) ∨ dove(Peter),
Christian Straßer (RUB, UGENT)
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Problems with Disjunctions
Let T = h∆, Φi where Quaker(x) : dove(x) republican(x) : hawk(x) , , ∆= dove(x) hawk(x) Φ = {Quaker(Peter) ∨ republican(Peter), Quaker(Anne) ∨ Quaker(George)}.
Problem we don’t get hawk(Peter) ∨ dove(Peter), dove(Anne) ∨ dove(George).
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"Naming defaults"
Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x)
Christian Straßer (RUB, UGENT)
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"Naming defaults"
Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of
Christian Straßer (RUB, UGENT)
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"Naming defaults"
Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of ∀x(birdsfly(x) ∧ bird(x) → flies(x))
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"Naming defaults"
Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of ∀x(birdsfly(x) ∧ bird(x) → flies(x)) ∀x(bird(x) ∧ baby(x) → ¬birdsfly(x))
Christian Straßer (RUB, UGENT)
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"Naming defaults"
Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of ∀x(birdsfly(x) ∧ bird(x) → flies(x)) ∀x(bird(x) ∧ baby(x) → ¬birdsfly(x)) bird(Tweety), bird(Polly)
Christian Straßer (RUB, UGENT)
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"Naming defaults"
Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of ∀x(birdsfly(x) ∧ bird(x) → flies(x)) ∀x(bird(x) ∧ baby(x) → ¬birdsfly(x)) bird(Tweety), bird(Polly) bird(Anne) ∨ bird(George)
Christian Straßer (RUB, UGENT)
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"Naming defaults"
Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of ∀x(birdsfly(x) ∧ bird(x) → flies(x)) ∀x(bird(x) ∧ baby(x) → ¬birdsfly(x)) bird(Tweety), bird(Polly) bird(Anne) ∨ bird(George) baby(Polly), baby(Keith)
Christian Straßer (RUB, UGENT)
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"Naming defaults"
Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of ∀x(birdsfly(x) ∧ bird(x) → flies(x)) ∀x(bird(x) ∧ baby(x) → ¬birdsfly(x)) bird(Tweety), bird(Polly) bird(Anne) ∨ bird(George) baby(Polly), baby(Keith) ¬flies(Fred)}
Christian Straßer (RUB, UGENT)
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"Naming defaults" The good: flies(Anne) ∨ flies(George) Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x)
flies(Tweety)
Φ consists of ∀x(birdsfly(x) ∧ bird(x) → flies(x)) ∀x(bird(x) ∧ baby(x) → ¬birdsfly(x)) bird(Tweety), bird(Polly) bird(Anne) ∨ bird(George) baby(Polly), baby(Keith) ¬flies(Fred)}
Christian Straßer (RUB, UGENT)
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"Naming defaults" The good: flies(Anne) ∨ flies(George) Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of
flies(Tweety)
The bad: But, in some respect this
∀x(birdsfly(x) ∧ bird(x) → flies(x)) ∀x(bird(x) ∧ baby(x) → ¬birdsfly(x)) proposal is too radical: bird(Tweety), bird(Polly) bird(Anne) ∨ bird(George) baby(Polly), baby(Keith) ¬flies(Fred)}
Christian Straßer (RUB, UGENT)
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"Naming defaults" The good: flies(Anne) ∨ flies(George) Let T = h∆, Φi where n o : birdsfly(x) ∆ = > birdsfly(x) Φ consists of
flies(Tweety)
The bad: But, in some respect this
∀x(birdsfly(x) ∧ bird(x) → flies(x)) ∀x(bird(x) ∧ baby(x) → ¬birdsfly(x)) proposal is too radical: bird(Tweety), bird(Polly) ¬bird(Keith) bird(Anne) ∨ bird(George) ¬bird(Fred) baby(Polly), baby(Keith) for any ground term ¬flies(Fred)}
t 6= Polly:
birdsfly(t) bird(t) → (flies(t) ∧ ¬baby(t)) Christian Straßer (RUB, UGENT)
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A different modelling of the same example Let T = h∆, Φi where
Christian Straßer (RUB, UGENT)
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A different modelling of the same example Let T = h∆, Φi where o n : birdsfly(x) ∆ = bird(x)birdsfly(x) Φ consists of ∀x(birdsfly(x) → flies(x)) ∀x(baby(x) → ¬birdsfly(x)) bird(Tweety), bird(Polly) bird(Anne) ∨ bird(George) baby(Polly), baby(Keith) ¬flies(Fred)}
Christian Straßer (RUB, UGENT)
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A different modelling of the same example Let T = h∆, Φi where o n : birdsfly(x) ∆ = bird(x)birdsfly(x) Φ consists of ∀x(birdsfly(x) → flies(x)) ∀x(baby(x) → ¬birdsfly(x)) bird(Tweety), bird(Polly) bird(Anne) ∨ bird(George) baby(Polly), baby(Keith) ¬flies(Fred)}
Task Try to see why some of the negative —too strong— consequences of the previous slide are avoided in this proposal. Try to see why flies(Anne) ∨ flies(George) is not anymore in any extension. Christian Straßer (RUB, UGENT)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x) Φ consists of
Christian Straßer (RUB, UGENT)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x) Φ consists of bird(Tweety), bird(Pete), bird(Mary)
Christian Straßer (RUB, UGENT)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x) Φ consists of bird(Tweety), bird(Pete), bird(Mary) baby(Polly)
Christian Straßer (RUB, UGENT)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x) Φ consists of bird(Tweety), bird(Pete), bird(Mary) baby(Polly) ¬flies(Fred)
Christian Straßer (RUB, UGENT)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x) Φ consists of bird(Tweety), bird(Pete), bird(Mary) baby(Polly) ¬flies(Fred) bird(Oscar) ∨ bird(Sylvester)
Christian Straßer (RUB, UGENT)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x) Φ consists of bird(Tweety), bird(Pete), bird(Mary) baby(Polly) ¬flies(Fred) bird(Oscar) ∨ bird(Sylvester) baby(Pete) ∨ baby(Mary)}
Christian Straßer (RUB, UGENT)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x) Φ consists of bird(Tweety), bird(Pete), bird(Mary) baby(Polly) ¬flies(Fred) bird(Oscar) ∨ bird(Sylvester) baby(Pete) ∨ baby(Mary)}
This prevents the conclusion ¬baby(Tweety).
Christian Straßer (RUB, UGENT)
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Semi-normal defaults Let T = h∆, Φi where n o flies(x)∧¬baby(x) ∆ = bird(x) : flies(x) Φ consists of bird(Tweety), bird(Pete), bird(Mary) baby(Polly) ¬flies(Fred) bird(Oscar) ∨ bird(Sylvester) baby(Pete) ∨ baby(Mary)}
This prevents the conclusion ¬baby(Tweety).
Task Try to see what problems we have with disjunctions for this example.
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The expressive power of semi-normal defaults Lukasziewicz writes: Assume, for instance, that on Sundays I usually go fishing, and suppose that you should remain agnostic about my fishing in rainy Sundays. It seems that the only appropriate representation of this situation is to use the following non-normal default: Sunday : I−go−fishing ∧ ¬rain I−go−fishing
Christian Straßer (RUB, UGENT)
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The expressive power of semi-normal defaults Lukasziewicz writes: Assume, for instance, that on Sundays I usually go fishing, and suppose that you should remain agnostic about my fishing in rainy Sundays. It seems that the only appropriate representation of this situation is to use the following non-normal default: Sunday : I−go−fishing ∧ ¬rain I−go−fishing
Critically evaluated this claim. 1 2
Why is a normal representation of this default suboptimal? Do you agree with L.’s assessment that the proposed non-normal representation is adequate? Suppose your only knowledge is that it is Sunday.
Christian Straßer (RUB, UGENT)
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Semi-normal defaults and the problem of inconsistent assumptions
Let T = h∆, Φi where bird(x) : flies(x) ∧ ¬dead(x) ∆= flies(x) of−ancient−species(x) : fossilised(x) ∧ dead(x) fossilised(x)
Christian Straßer (RUB, UGENT)
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Semi-normal defaults and the problem of inconsistent assumptions
Let T = h∆, Φi where bird(x) : flies(x) ∧ ¬dead(x) ∆= flies(x) of−ancient−species(x) : fossilised(x) ∧ dead(x) fossilised(x) Φ = {bird(Tweety), of−ancient−species(Tweety)}
Christian Straßer (RUB, UGENT)
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Semi-normal defaults and the problem of inconsistent assumptions
Let T = h∆, Φi where bird(x) : flies(x) ∧ ¬dead(x) ∆= flies(x) of−ancient−species(x) : fossilised(x) ∧ dead(x) fossilised(x) Φ = {bird(Tweety), of−ancient−species(Tweety)}
Task Try to see what’s the problem here.
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Semi-normal vs. normal defaults
Compare has−motive(x) : guilty(x) ∧ suspect(x) suspect(x) with has−motive(x) : guilty(x) ∧ suspect(x) guilty(x) ∧ suspect(x)
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Semi-Normal defaults ala Brewka/Levesque Let T = h∆, Φi where n o > : flies(x) ∆ = bird(x)→flies(x)
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Semi-Normal defaults ala Brewka/Levesque Let T = h∆, Φi where n o > : flies(x) ∆ = bird(x)→flies(x) Φ consists of
Christian Straßer (RUB, UGENT)
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Semi-Normal defaults ala Brewka/Levesque Let T = h∆, Φi where n o > : flies(x) ∆ = bird(x)→flies(x) Φ consists of bird(Tweety), bird(Polly)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Semi-Normal defaults ala Brewka/Levesque Let T = h∆, Φi where n o > : flies(x) ∆ = bird(x)→flies(x) Φ consists of bird(Tweety), bird(Polly) baby(Polly), ¬flies(Fred)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Semi-Normal defaults ala Brewka/Levesque Let T = h∆, Φi where n o > : flies(x) ∆ = bird(x)→flies(x) Φ consists of bird(Tweety), bird(Polly) baby(Polly), ¬flies(Fred) bird(Oscar) ∨ bird(Sylvester)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Semi-Normal defaults ala Brewka/Levesque Let T = h∆, Φi where n o > : flies(x) ∆ = bird(x)→flies(x) Φ consists of bird(Tweety), bird(Polly) baby(Polly), ¬flies(Fred) bird(Oscar) ∨ bird(Sylvester) ¬flies(Pete) ∨ ¬flies(Mary)}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Semi-Normal defaults ala Brewka/Levesque Let T = h∆, Φi where n o > : flies(x) ∆ = bird(x)→flies(x) Φ consists of bird(Tweety), bird(Polly) baby(Polly), ¬flies(Fred) bird(Oscar) ∨ bird(Sylvester) ¬flies(Pete) ∨ ¬flies(Mary)}
Task We cannot conclude ¬bird(Fred). See why.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Semi-Normal defaults ala Brewka/Levesque Let T = h∆, Φi where n o > : flies(x) ∆ = bird(x)→flies(x) Φ consists of bird(Tweety), bird(Polly) baby(Polly), ¬flies(Fred) bird(Oscar) ∨ bird(Sylvester) ¬flies(Pete) ∨ ¬flies(Mary)}
Task We cannot conclude ¬bird(Fred). See why. Observe what happens to Pete and Mary.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The Finch-Example
Let T = h∆, n Φi where : green−island(x) ∆ = ruffed−finch(x) , green−island(x) least−ruffed−finch(x) : green−island(x)∨sand−island(x) green−island(x)∨sand−island(x)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
o
September 2, 2015
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The Finch-Example
Let T = h∆, n Φi where : green−island(x) ∆ = ruffed−finch(x) , green−island(x) least−ruffed−finch(x) : green−island(x)∨sand−island(x) green−island(x)∨sand−island(x)
o
Φ consists of
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The Finch-Example
Let T = h∆, n Φi where : green−island(x) ∆ = ruffed−finch(x) , green−island(x) least−ruffed−finch(x) : green−island(x)∨sand−island(x) green−island(x)∨sand−island(x)
o
Φ consists of least−ruffed−finch(Frank)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The Finch-Example
Let T = h∆, n Φi where : green−island(x) ∆ = ruffed−finch(x) , green−island(x) least−ruffed−finch(x) : green−island(x)∨sand−island(x) green−island(x)∨sand−island(x)
o
Φ consists of least−ruffed−finch(Frank) ∀x(least−ruffed−finch(x) → ruffed−finch(x))}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The Finch-Example
Let T = h∆, n Φi where : green−island(x) ∆ = ruffed−finch(x) , green−island(x) least−ruffed−finch(x) : green−island(x)∨sand−island(x) green−island(x)∨sand−island(x)
o
Φ consists of least−ruffed−finch(Frank) ∀x(least−ruffed−finch(x) → ruffed−finch(x))}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The Finch-Example
Let T = h∆, n Φi where : green−island(x) ∆ = ruffed−finch(x) , green−island(x) least−ruffed−finch(x) : green−island(x)∨sand−island(x) green−island(x)∨sand−island(x)
o
Φ consists of least−ruffed−finch(Frank) ∀x(least−ruffed−finch(x) → ruffed−finch(x))}
Problem
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The Finch-Example
Let T = h∆, n Φi where : green−island(x) ∆ = ruffed−finch(x) , green−island(x) least−ruffed−finch(x) : green−island(x)∨sand−island(x) green−island(x)∨sand−island(x)
o
Φ consists of least−ruffed−finch(Frank) ∀x(least−ruffed−finch(x) → ruffed−finch(x))}
Problem the unique extension includes both green−island(Frank) and green−island(Frank) ∨ sand−island(Frank) (since both defaults are triggered) Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Default logic and monotonicity Nonmonotonicity, both in the set of defaults ∆ in the set of facts Φ
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Default logic and monotonicity Nonmonotonicity, both in the set of defaults ∆ in the set of facts Φ
Not even cautious monotonic Here’s an example that goes back to Makinson:
∆=
n
> : p p∨q : ¬p , p ¬p
o
Φ1 = ∅ Φ2 = {p ∨ q} Check, what happens! Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The problem with ’negative cycles’
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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The problem with ’negative cycles’
for a way to deal with this problem see Antonelli (1999)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Prioritized default theories (Horty (2007, 2012))
hΦ, ∆, ≺i where we have a strict partial order on the defaults: ≺. This represents a priority/preference relation. Depending on the application this may indicate the rank of an authority from which the information stems, the reliability of the source, specificity relations, etc.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Prioritized default theories (Horty (2007, 2012))
hΦ, ∆, ≺i where we have a strict partial order on the defaults: ≺. This represents a priority/preference relation. Depending on the application this may indicate the rank of an authority from which the information stems, the reliability of the source, specificity relations, etc. A strict partial order is (i) irreflexive (A 6< A), (ii) asymmetric (If A < B then B 6< A), and (iii) transitive (A < B and B < C implies A < C ). Graphically they are represented by directed acyclic graphs (the transitive closure is usually not represented).
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Scenarios vs. extensions
Given an ordered default theory hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Scenarios vs. extensions
Given an ordered default theory hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Proper Scenarios
like in the non-prioritised case, not just any superset of Φ constitutes an extension, we are also now interested in scenarios that in some sense represent the set of defaults a rational agent would select/use given Φ and <.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Proper Scenarios
like in the non-prioritised case, not just any superset of Φ constitutes an extension, we are also now interested in scenarios that in some sense represent the set of defaults a rational agent would select/use given Φ and <. E.g., a scenario should not generate conflicting beliefs.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Proper Scenarios
like in the non-prioritised case, not just any superset of Φ constitutes an extension, we are also now interested in scenarios that in some sense represent the set of defaults a rational agent would select/use given Φ and <. E.g., a scenario should not generate conflicting beliefs. Moreover, priorities should be taken into account.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Proper Scenarios
like in the non-prioritised case, not just any superset of Φ constitutes an extension, we are also now interested in scenarios that in some sense represent the set of defaults a rational agent would select/use given Φ and <. E.g., a scenario should not generate conflicting beliefs. Moreover, priorities should be taken into account. Hence, we are interested in what Horty calls proper scenarios and the belief sets generated by them.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The idea is again to build up scenarios stepwise similar as in the procedural approaches to build extensions.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The idea is again to build up scenarios stepwise similar as in the procedural approaches to build extensions. we start with our facts Φ
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The idea is again to build up scenarios stepwise similar as in the procedural approaches to build extensions. we start with our facts Φ What are interesting defaults to take into account?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The idea is again to build up scenarios stepwise similar as in the procedural approaches to build extensions. we start with our facts Φ What are interesting defaults to take into account? we have the choice between triggered defaults: Triggered(S) = {δ ∈ ∆ | Φ ∪ Conclusion(S) ` Premise(δ)}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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The idea is again to build up scenarios stepwise similar as in the procedural approaches to build extensions. we start with our facts Φ What are interesting defaults to take into account? we have the choice between triggered defaults: Triggered(S) = {δ ∈ ∆ | Φ ∪ Conclusion(S) ` Premise(δ)} However, some triggered defaults may bad candidates since they (i.e., their conclusions) conflict with our belief set. Hence, we want to neglect: Conflicted(S) = {δ ∈ ∆ | Φ ∪ Conclusion(S) ` ¬Conclusion(δ)}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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What about the priorities? Also, we have to take into account our priorities.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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What about the priorities? Also, we have to take into account our priorities. A first idea would be to pick one of the highest ranked triggered defaults.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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What about the priorities? Also, we have to take into account our priorities. A first idea would be to pick one of the highest ranked triggered defaults. this is not Horty’s approach (see later: the order puzzle)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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What about the priorities? Also, we have to take into account our priorities. A first idea would be to pick one of the highest ranked triggered defaults. this is not Horty’s approach (see later: the order puzzle)
We also have to take into account that sometimes sets of defaults “defeat” other defaults
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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What about the priorities? Also, we have to take into account our priorities. A first idea would be to pick one of the highest ranked triggered defaults. this is not Horty’s approach (see later: the order puzzle)
We also have to take into account that sometimes sets of defaults “defeat” other defaults We write S < S 0 where S, S 0 ⊆ ∆ and δ < δ 0 for all δ ∈ S and all δ0 ∈ S 0
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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What about the priorities? Also, we have to take into account our priorities. A first idea would be to pick one of the highest ranked triggered defaults. this is not Horty’s approach (see later: the order puzzle)
We also have to take into account that sometimes sets of defaults “defeat” other defaults We write S < S 0 where S, S 0 ⊆ ∆ and δ < δ 0 for all δ ∈ S and all δ0 ∈ S 0
Example E.g., where δ1 = a → b, δ2 = b → c and δ3 = a → ¬c and δ3 < δ1 , δ2 , we have {δ3 } < {δ1 , δ2 } Note that {a} ∪ Conclusion({δ1 , δ2 }) ` ¬Conclusion(δ3 ) in this sense {δ1 , δ2 } defeats δ3 Christian Straßer (RUB, UGENT)
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Horty makes the idea above precise relative to a given scenario S.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Horty makes the idea above precise relative to a given scenario S. The set Defeated(S) is the set of all δ ∈ ∆ such that there is a ∆d ⊆ Triggered(S) (a defeating set) for which
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty makes the idea above precise relative to a given scenario S. The set Defeated(S) is the set of all δ ∈ ∆ such that there is a ∆d ⊆ Triggered(S) (a defeating set) for which {δ} < ∆d and
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty makes the idea above precise relative to a given scenario S. The set Defeated(S) is the set of all δ ∈ ∆ such that there is a ∆d ⊆ Triggered(S) (a defeating set) for which {δ} < ∆d and there is a Sa ⊆ S (an accommodation set) for which
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty makes the idea above precise relative to a given scenario S. The set Defeated(S) is the set of all δ ∈ ∆ such that there is a ∆d ⊆ Triggered(S) (a defeating set) for which {δ} < ∆d and there is a Sa ⊆ S (an accommodation set) for which 1
Sa < ∆d
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty makes the idea above precise relative to a given scenario S. The set Defeated(S) is the set of all δ ∈ ∆ such that there is a ∆d ⊆ Triggered(S) (a defeating set) for which {δ} < ∆d and there is a Sa ⊆ S (an accommodation set) for which 1 2
Sa < ∆d Φ ∪ Conclusion((S \ Sa ) ∪ ∆d ) is consistent
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty makes the idea above precise relative to a given scenario S. The set Defeated(S) is the set of all δ ∈ ∆ such that there is a ∆d ⊆ Triggered(S) (a defeating set) for which {δ} < ∆d and there is a Sa ⊆ S (an accommodation set) for which 1 2 3
Sa < ∆d Φ ∪ Conclusion((S \ Sa ) ∪ ∆d ) is consistent Φ ∪ Conclusion((S \ Sa ) ∪ ∆d ) ` ¬Conclusion(δ)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Some examples
Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Suppose S = {B → F }.
Christian Straßer (RUB, UGENT)
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Some examples
Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Suppose S = {B → F }. P → ¬F ∈ Triggered(S)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Some examples
Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Suppose S = {B → F }. P → ¬F ∈ Triggered(S) {B → F } is an accommodation set for the defeating set {P → ¬F } w.r.t. B → F
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Let S1 = {a → ¬c}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Let S1 = {a → ¬c} a → b is triggered by S
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Let S1 = {a → ¬c} a → b is triggered by S no defeating takes place
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Let S1 = {a → ¬c} a → b is triggered by S no defeating takes place
Let S2 = {a → ¬c, a → b}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Let S1 = {a → ¬c} a → b is triggered by S no defeating takes place
Let S2 = {a → ¬c, a → b} b → c is triggered
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Let S1 = {a → ¬c} a → b is triggered by S no defeating takes place
Let S2 = {a → ¬c, a → b} b → c is triggered b → c is conflicted by S
Christian Straßer (RUB, UGENT)
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September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Let S1 = {a → ¬c} a → b is triggered by S no defeating takes place
Let S2 = {a → ¬c, a → b} b → c is triggered b → c is conflicted by S {a → ¬c} is defeated by S
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Let S1 = {a → ¬c} a → b is triggered by S no defeating takes place
Let S2 = {a → ¬c, a → b} b → c is triggered b → c is conflicted by S {a → ¬c} is defeated by S note that {a → ¬c} is an accommodation set for the defeating set {b → c} Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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An intuitive problem?
Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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An intuitive problem?
Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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An intuitive problem?
Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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An intuitive problem?
Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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An intuitive problem?
Take T = hΦ, ∆,
Φ = {a}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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An intuitive problem?
Take T = hΦ, ∆,
Φ = {a} δ1 < δ3
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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An intuitive problem?
Take T = hΦ, ∆,
Φ = {a} δ1 < δ3
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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An intuitive problem?
Take T = hΦ, ∆,
Φ = {a} δ1 < δ3
Take S = {a → b, b → c}
Note that a → b ∈ Defeated(S) {a → b} is an accommodation set for the defeating set {c → ¬b} Note that Φ ∪ Conclusion(b → c, c → ¬b) ` ¬b However, neither b → c nor c → ¬b is triggered w.r.t. Φ!
Christian Straßer (RUB, UGENT)
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S
Christian Straßer (RUB, UGENT)
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S
2
the initial scenario is S0 = ∅
Christian Straßer (RUB, UGENT)
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September 2, 2015
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S the initial scenario is S0 = ∅ do the following until a fixed point is reached
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S the initial scenario is S0 = ∅ do the following until a fixed point is reached 1
Si+1 : add all δ ∈ ∆ to Si that satisfy the following conditions
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S the initial scenario is S0 = ∅ do the following until a fixed point is reached 1
Si+1 : add all δ ∈ ∆ to Si that satisfy the following conditions 1
δ ∈ Triggered(Si )
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S the initial scenario is S0 = ∅ do the following until a fixed point is reached 1
Si+1 : add all δ ∈ ∆ to Si that satisfy the following conditions 1 2
δ ∈ Triggered(Si ) δ∈ / Conflicted(S) (here you need to make use of your guess!)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S the initial scenario is S0 = ∅ do the following until a fixed point is reached 1
Si+1 : add all δ ∈ ∆ to Si that satisfy the following conditions 1 2 3
δ ∈ Triggered(Si ) δ∈ / Conflicted(S) (here you need to make use of your guess!) δ∈ / Defeated(S) (also here)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S the initial scenario is S0 = ∅ do the following until a fixed point is reached 1
Si+1 : add all δ ∈ ∆ to Si that satisfy the following conditions 1 2 3
4
δ ∈ Triggered(Si ) δ∈ / Conflicted(S) (here you need to make use of your guess!) δ∈ / Defeated(S) (also here)
let your fixed point be S 0 .
Christian Straßer (RUB, UGENT)
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S the initial scenario is S0 = ∅ do the following until a fixed point is reached 1
Si+1 : add all δ ∈ ∆ to Si that satisfy the following conditions 1 2 3
4
δ ∈ Triggered(Si ) δ∈ / Conflicted(S) (here you need to make use of your guess!) δ∈ / Defeated(S) (also here)
let your fixed point be S 0 . 1
If S = S 0 you’re done. Then S is a proper scenario and Ξ = Cn(Φ ∪ Conclusion(S)) is an extension.
Christian Straßer (RUB, UGENT)
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September 2, 2015
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Horty’s procedure for proper scenarios and extensions Given an ordered default theory hΦ, ∆,
guess S the initial scenario is S0 = ∅ do the following until a fixed point is reached 1
Si+1 : add all δ ∈ ∆ to Si that satisfy the following conditions 1 2 3
4
δ ∈ Triggered(Si ) δ∈ / Conflicted(S) (here you need to make use of your guess!) δ∈ / Defeated(S) (also here)
let your fixed point be S 0 . 1
2
If S = S 0 you’re done. Then S is a proper scenario and Ξ = Cn(Φ ∪ Conclusion(S)) is an extension. Otherwise, start anew with another guess.
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }.
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round: S0 = ∅
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 )
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S)
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S)
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Guess: S = {P → ¬F }
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Guess: S = {P → ¬F } First round:
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Guess: S = {P → ¬F } First round: S0 = ∅
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Guess: S = {P → ¬F } First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 )
Christian Straßer (RUB, UGENT)
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September 2, 2015
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Guess: S = {P → ¬F } First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) B → F ∈ Conflicted(S)
Christian Straßer (RUB, UGENT)
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September 2, 2015
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Guess: S = {P → ¬F } First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) B → F ∈ Conflicted(S) P → ¬F ∈ / Defeated(S)
Christian Straßer (RUB, UGENT)
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September 2, 2015
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Guess: S = {P → ¬F } First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) B → F ∈ Conflicted(S) P → ¬F ∈ / Defeated(S) hence: S1 = {P → ¬F }
Christian Straßer (RUB, UGENT)
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An example Take T = h{P, P ⊃ B}, {δ1 , δ2 }, {(δ1 , δ2 )}i where δ1 = B → F and δ2 = P → ¬F . Guess S = {B → F }. First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) P → ¬F ∈ Conflicted(S) but: B → F ∈ Defeated(S) nothing is added!
Guess: S = {P → ¬F } First round:
S0 = ∅ B → F , P → ¬F ∈ Triggered(S0 ) B → F ∈ Conflicted(S) P → ¬F ∈ / Defeated(S) hence: S1 = {P → ¬F }
second round: fixed point reached. Christian Straßer (RUB, UGENT)
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Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Christian Straßer (RUB, UGENT)
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Take T = hΦ, ∆,
Φ = {a}
Christian Straßer (RUB, UGENT)
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2
Christian Straßer (RUB, UGENT)
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Guess: S = {a → ¬c}
Christian Straßer (RUB, UGENT)
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Guess: S = {a → ¬c} Round 1: S0 = ∅
Christian Straßer (RUB, UGENT)
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Guess: S = {a → ¬c} Round 1: S0 = ∅ Triggered: a → b and a → ¬c
Christian Straßer (RUB, UGENT)
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Guess: S = {a → ¬c} Round 1: S0 = ∅ Triggered: a → b and a → ¬c a→b∈ / Defeated(S)
Christian Straßer (RUB, UGENT)
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Guess: S = {a → ¬c} Round 1: S0 = ∅ Triggered: a → b and a → ¬c a→b∈ / Defeated(S) a→b∈ / Conflicted(S)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Guess: S = {a → ¬c} Round 1: S0 = ∅ Triggered: a → b and a → ¬c a→b∈ / Defeated(S) a→b∈ / Conflicted(S) hence, a → b has to be added!
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Guess: S = {a → ¬c} Round 1: S0 = ∅ Triggered: a → b and a → ¬c a→b∈ / Defeated(S) a→b∈ / Conflicted(S) hence, a → b has to be added!
Guess: S = {a → b, b → c}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Take T = hΦ, ∆,
Φ = {a} δ3 < δ1 and δ3 < δ2 Guess: S = {a → ¬c} Round 1: S0 = ∅ Triggered: a → b and a → ¬c a→b∈ / Defeated(S) a→b∈ / Conflicted(S) hence, a → b has to be added!
Guess: S = {a → b, b → c} This works: check out why
Christian Straßer (RUB, UGENT)
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Theorem Where hΦ, ∆,
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Theorem Where hΦ, ∆,
Theorem Where T< = hΦ, ∆,
Christian Straßer (RUB, UGENT)
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Another problematic example: The Order Puzzle Let T = hΦ, ∆, → A δ2 = > → B δ3 = A → ¬B
Φ=∅ δ1 < δ2 < δ3
Christian Straßer (RUB, UGENT)
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Another problematic example: The Order Puzzle Let T = hΦ, ∆, → A δ2 = > → B δ3 = A → ¬B
Φ=∅ δ1 < δ2 < δ3
We have one proper scenarios:
Christian Straßer (RUB, UGENT)
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September 2, 2015
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Another problematic example: The Order Puzzle Let T = hΦ, ∆, → A δ2 = > → B δ3 = A → ¬B
Φ=∅ δ1 < δ2 < δ3
We have one proper scenarios: S1 = {δ1 , δ2 }? Nope!
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Another problematic example: The Order Puzzle Let T = hΦ, ∆, → A δ2 = > → B δ3 = A → ¬B
Φ=∅ δ1 < δ2 < δ3
We have one proper scenarios: S1 = {δ1 , δ2 }? Nope! S2 = {δ1 , δ3 }
What do you think? Interpret the premises in terms of conditional commands, and in terms of various pieces of information with different degrees of reliability. Christian Straßer (RUB, UGENT)
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The normative reading by Horty (p.392, 2007)
Christian Straßer (RUB, UGENT)
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A similar problem Let RC = “being a resident of Cuba”; RN = “being a resident of North-America; CC = “being a citizen of Cuba”; CU = “being a citizen of the USA”; and VU = “having voting rights in the USA”. Let T = hΦ, ∆,
Φ = {RC , RC ⊃ RN, ¬(CC ∧ CU), ¬(CC ∧ VU)} δ1 < δ2 < δ3
Christian Straßer (RUB, UGENT)
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A similar problem Let RC = “being a resident of Cuba”; RN = “being a resident of North-America; CC = “being a citizen of Cuba”; CU = “being a citizen of the USA”; and VU = “having voting rights in the USA”. Let T = hΦ, ∆,
Φ = {RC , RC ⊃ RN, ¬(CC ∧ CU), ¬(CC ∧ VU)} δ1 < δ2 < δ3
There are two proper scenarios: S1 = {δ2 } and S2 = {δ1 , δ3 }. (this is problematic) Christian Straßer (RUB, UGENT)
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September 2, 2015
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An intuitive problem?
Take T = hΦ, ∆,
Φ = {a} δ1 < δ3
Christian Straßer (RUB, UGENT)
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September 2, 2015
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An intuitive problem?
Take T = hΦ, ∆,
Φ = {a} δ1 < δ3
No extension/proper scenario!
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Multiple occurrences of the same defaults Let T = hΦ, ∆,
=>→A = > → ¬A =>→A = > → ¬A
Φ=∅ δ1 < δ2 and δ3 < δ4
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Multiple occurrences of the same defaults Let T = hΦ, ∆,
=>→A = > → ¬A =>→A = > → ¬A
Φ=∅ δ1 < δ2 and δ3 < δ4
We have two proper scenarios: S1 = {δ2 , δ4 } S2 = {δ1 , δ3 } (problematic!)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Multiple occurrences of the same defaults Let T = hΦ, ∆,
=>→A = > → ¬A =>→A = > → ¬A
Φ=∅ δ1 < δ2 and δ3 < δ4
We have two proper scenarios: S1 = {δ2 , δ4 } S2 = {δ1 , δ3 } (problematic!)
Solution . . . redefine < for sets of defaults? Christian Straßer (RUB, UGENT)
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Some philosophical Questions in the end
What should we derive in default logic? What is our standard of adequacy, what is our ideal rational agent?
Christian Straßer (RUB, UGENT)
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Some philosophical Questions in the end
What should we derive in default logic? What is our standard of adequacy, what is our ideal rational agent? How descriptive of actual reasoning should default logic be, and how normative should it be?
Christian Straßer (RUB, UGENT)
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September 2, 2015
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Some philosophical Questions in the end
What should we derive in default logic? What is our standard of adequacy, what is our ideal rational agent? How descriptive of actual reasoning should default logic be, and how normative should it be? What are the “core properties” of default reasoning?
Christian Straßer (RUB, UGENT)
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Looking back . . . : what did we learn?
What is a default theory?
Christian Straßer (RUB, UGENT)
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Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions.
Christian Straßer (RUB, UGENT)
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September 2, 2015
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Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions. There may be multiple extensions, for Reiter’s logic there may be none.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions. There may be multiple extensions, for Reiter’s logic there may be none. We have highlighted various problems for default logic:
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions. There may be multiple extensions, for Reiter’s logic there may be none. We have highlighted various problems for default logic: specificity
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions. There may be multiple extensions, for Reiter’s logic there may be none. We have highlighted various problems for default logic: specificity disjunctive facts
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions. There may be multiple extensions, for Reiter’s logic there may be none. We have highlighted various problems for default logic: specificity disjunctive facts disjunctive exceptions
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
93 / 104
Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions. There may be multiple extensions, for Reiter’s logic there may be none. We have highlighted various problems for default logic: specificity disjunctive facts disjunctive exceptions consistency of assumptions
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
93 / 104
Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions. There may be multiple extensions, for Reiter’s logic there may be none. We have highlighted various problems for default logic: specificity disjunctive facts disjunctive exceptions consistency of assumptions etc.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Looking back . . . : what did we learn?
What is a default theory? We learned about two different notions for extensions. There may be multiple extensions, for Reiter’s logic there may be none. We have highlighted various problems for default logic: specificity disjunctive facts disjunctive exceptions consistency of assumptions etc.
There are various ways of phrasing defaults (normal vs. semi-normal vs. naming defaults, etc.)
Christian Straßer (RUB, UGENT)
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Topic 1
2 3
4 5
Defeasible Reasoning – Some Basic Concepts Some examples to warm up Some conceptual distinctions Nonmonotonic Logic in Context The Dynamics of Defeasible Reasoning Default Logic (in the tradition of Reiter) Warming up Defaults and Default Theories Inferring Alternatives and more examples Meta-Properties Introducing Priorities Summing up Autoepistemic Logic Bibliography Bibliography
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Idea (Moore (1984, 1985))
Logic to model the reasoning of an (ideal) agent reflecting about her beliefs
Christian Straßer (RUB, UGENT)
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Idea (Moore (1984, 1985))
Logic to model the reasoning of an (ideal) agent reflecting about her beliefs
Autoepistemic Theory features the beliefs A of an epistemic agent
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Idea (Moore (1984, 1985))
Logic to model the reasoning of an (ideal) agent reflecting about her beliefs
Autoepistemic Theory features the beliefs A of an epistemic agent including her introspective beliefs about her beliefs BA (and lack thereof)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Idea (Moore (1984, 1985))
Logic to model the reasoning of an (ideal) agent reflecting about her beliefs
Autoepistemic Theory features the beliefs A of an epistemic agent including her introspective beliefs about her beliefs BA (and lack thereof) closed under rationality constraints (next slide!)
Christian Straßer (RUB, UGENT)
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Stable Autoepistemic Theories
Stalnaker (1993), stability criteria on a autoepistemic theory Γ 1
A ∈ Γ implies BA ∈ Γ
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Stable Autoepistemic Theories
Stalnaker (1993), stability criteria on a autoepistemic theory Γ 1
A ∈ Γ implies BA ∈ Γ
2
A∈ / Γ implies ¬BA ∈ Γ.
Christian Straßer (RUB, UGENT)
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Stable Autoepistemic Theories
Stalnaker (1993), stability criteria on a autoepistemic theory Γ 1
A ∈ Γ implies BA ∈ Γ
2
A∈ / Γ implies ¬BA ∈ Γ.
3
classical closure: Cn(Γ) = Γ.
Christian Straßer (RUB, UGENT)
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Stable Autoepistemic Theories
Stalnaker (1993), stability criteria on a autoepistemic theory Γ 1
A ∈ Γ implies BA ∈ Γ
2
A∈ / Γ implies ¬BA ∈ Γ.
3
classical closure: Cn(Γ) = Γ.
If Γ is consistent, this implies . . .
Christian Straßer (RUB, UGENT)
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Stable Autoepistemic Theories
Stalnaker (1993), stability criteria on a autoepistemic theory Γ 1
A ∈ Γ implies BA ∈ Γ
2
A∈ / Γ implies ¬BA ∈ Γ.
3
classical closure: Cn(Γ) = Γ.
If Γ is consistent, this implies . . . 1
A ∈ Γ iff BA ∈ Γ
2
A∈ / Γ iff ¬BA.
You see why?
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Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1 by cons., brother ∈ / Ξ1
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1 by cons., brother ∈ / Ξ1 by (2), ¬Bbrother ∈ Ξ1
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1 by cons., brother ∈ / Ξ1 by (2), ¬Bbrother ∈ Ξ1 by (1), Bi ¬Bbrother ∈ Ξ1
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1
Ξ2 containing brother Can you see what else is contained?
by cons., brother ∈ / Ξ1 by (2), ¬Bbrother ∈ Ξ1 by (1), Bi ¬Bbrother ∈ Ξ1
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1 by cons., brother ∈ / Ξ1
Ξ2 containing brother Can you see what else is contained? by (1), Bi brother ∈ Ξ2
by (2), ¬Bbrother ∈ Ξ1 by (1), Bi ¬Bbrother ∈ Ξ1
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1 by cons., brother ∈ / Ξ1 by (2), ¬Bbrother ∈ Ξ1
Ξ2 containing brother Can you see what else is contained? by (1), Bi brother ∈ Ξ2 by cons., ¬brother ∈ / Ξ2
by (1), Bi ¬Bbrother ∈ Ξ1
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1 by cons., brother ∈ / Ξ1 by (2), ¬Bbrother ∈ Ξ1 by (1), Bi ¬Bbrother ∈ Ξ1
Christian Straßer (RUB, UGENT)
Ξ2 containing brother Can you see what else is contained? by (1), Bi brother ∈ Ξ2 by cons., ¬brother ∈ / Ξ2 by (2), ¬B¬brother ∈ Ξ2
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1 by cons., brother ∈ / Ξ1 by (2), ¬Bbrother ∈ Ξ1 by (1), Bi ¬Bbrother ∈ Ξ1
Ξ2 containing brother Can you see what else is contained? by (1), Bi brother ∈ Ξ2 by cons., ¬brother ∈ / Ξ2 by (2), ¬B¬brother ∈ Ξ2
Do you see a problem?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
97 / 104
Stability is not enough . . . Groundedness Take Ξ = {¬Bbrother ⊃ ¬brother}.
Two consistent stable autoepistemic supersets of Ξ: do you see which? Ξ1 containing ¬brother Can you see what else is contained? by (1), Bi ¬brother ∈ Ξ1 by cons., brother ∈ / Ξ1 by (2), ¬Bbrother ∈ Ξ1 by (1), Bi ¬Bbrother ∈ Ξ1
Ξ2 containing brother Can you see what else is contained? by (1), Bi brother ∈ Ξ2 by cons., ¬brother ∈ / Ξ2 by (2), ¬B¬brother ∈ Ξ2
Do you see a problem? The belief in brother in Ξ2 is not grounded given Ξ! Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Grounding Autoepistemic Theories Here’s how it’s done:
Autoepistemic Extension of Γ Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ}) where Ξ = {A | A ∈ / Γ} BΞ = {BA | A ∈ Ξ}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Grounding Autoepistemic Theories Here’s how it’s done:
Autoepistemic Extension of Γ Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ}) where Ξ = {A | A ∈ / Γ} BΞ = {BA | A ∈ Ξ} Do you see where there is no autoepistemic extension of Γ = {¬Bbrother ⊃ ¬brother} that contains brother?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
98 / 104
Grounding Autoepistemic Theories Here’s how it’s done:
Autoepistemic Extension of Γ Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ}) where Ξ = {A | A ∈ / Γ} BΞ = {BA | A ∈ Ξ} Do you see where there is no autoepistemic extension of Γ = {¬Bbrother ⊃ ¬brother} that contains brother? Suppose there were one Ξ! But then Γ ∪ BΞ ∪ ¬BΞ ` brother — but this is impossible!
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ:
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ: but how to ground A?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ: but how to ground A? in view of Γ the only way is to have also ¬BA
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ: but how to ground A? in view of Γ the only way is to have also ¬BA by we have BA ∈ Ξ
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ: but how to ground A? in view of Γ the only way is to have also ¬BA by we have BA ∈ Ξ hence, Ξ is trivial (by classical closure)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ: but how to ground A? in view of Γ the only way is to have also ¬BA by we have BA ∈ Ξ hence, Ξ is trivial (by classical closure) but then BΞ = ∅, a contradiction, since now there is no way to derive ¬BA
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ: but how to ground A? in view of Γ the only way is to have also ¬BA by we have BA ∈ Ξ hence, Ξ is trivial (by classical closure) but then BΞ = ∅, a contradiction, since now there is no way to derive ¬BA
suppose A ∈ / Ξ:
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ: but how to ground A? in view of Γ the only way is to have also ¬BA by we have BA ∈ Ξ hence, Ξ is trivial (by classical closure) but then BΞ = ∅, a contradiction, since now there is no way to derive ¬BA
suppose A ∈ / Ξ: then ¬BA
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Extension do not always exist Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Take Γ = {¬BA ⊃ A}. Let Ξ be any classically closed superset of Γ. suppose A ∈ Ξ: but how to ground A? in view of Γ the only way is to have also ¬BA by we have BA ∈ Ξ hence, Ξ is trivial (by classical closure) but then BΞ = ∅, a contradiction, since now there is no way to derive ¬BA
suppose A ∈ / Ξ: then ¬BA but then also A by modus ponens, contradiction Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
99 / 104
Are autoepistemic extensions of some Γ always consistent? Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
100 / 104
Are autoepistemic extensions of some Γ always consistent? Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
If Γ is inconsistent: nope!
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Are autoepistemic extensions of some Γ always consistent? Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
If Γ is inconsistent: nope! What if Γ is consistent?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
100 / 104
Are autoepistemic extensions of some Γ always consistent? Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
If Γ is inconsistent: nope! What if Γ is consistent? Take Γ = {¬Bp} if p is in we get Bp together with Γ we can derive anything
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q and ¬Bq to ground p
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q and ¬Bq to ground p
2
what about one with ¬p and/or ¬q?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q and ¬Bq to ground p
2
what about one with ¬p and/or ¬q? again: how to ground ¬p and/or ¬q?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q and ¬Bq to ground p
2
what about one with ¬p and/or ¬q? again: how to ground ¬p and/or ¬q? only possibility is if Γ were inconsistent: but it isn’t
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q and ¬Bq to ground p
2
what about one with ¬p and/or ¬q? again: how to ground ¬p and/or ¬q? only possibility is if Γ were inconsistent: but it isn’t
3
what about one with p and without q?
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q and ¬Bq to ground p
2
what about one with ¬p and/or ¬q? again: how to ground ¬p and/or ¬q? only possibility is if Γ were inconsistent: but it isn’t
3
what about one with p and without q? then ¬Bq (neg. introspection)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q and ¬Bq to ground p
2
what about one with ¬p and/or ¬q? again: how to ground ¬p and/or ¬q? only possibility is if Γ were inconsistent: but it isn’t
3
what about one with p and without q? then ¬Bq (neg. introspection) thus, p is grounded
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Another Example Take Γ = {¬Bp ⊃ q, ¬Bq ⊃ p}. What are the autoepistemic extensions of Γ? 1
what about one with p and q? but how to ground p and q? We would need ¬Bp to ground q and ¬Bq to ground p
2
what about one with ¬p and/or ¬q? again: how to ground ¬p and/or ¬q? only possibility is if Γ were inconsistent: but it isn’t
3
what about one with p and without q? then ¬Bq (neg. introspection) thus, p is grounded
4
similarly, one with q and without p
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
101 / 104
Bootstrapping problem (Konolige (1988)) Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
The problem Take Γ = {BA ⊃ A}.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
102 / 104
Bootstrapping problem (Konolige (1988)) Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
The problem Take Γ = {BA ⊃ A}. There is an autoepistemic extension that contains A (and BA and A ∨ B and B(A ∨ B) and ¬B¬A, etc.). Note that A is grounded in view of Γ once we adopt BA.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
102 / 104
Bootstrapping problem (Konolige (1988)) Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
The problem Take Γ = {BA ⊃ A}. There is an autoepistemic extension that contains A (and BA and A ∨ B and B(A ∨ B) and ¬B¬A, etc.). Note that A is grounded in view of Γ once we adopt BA.
Solution
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
102 / 104
Bootstrapping problem (Konolige (1988)) Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
The problem Take Γ = {BA ⊃ A}. There is an autoepistemic extension that contains A (and BA and A ∨ B and B(A ∨ B) and ¬B¬A, etc.). Note that A is grounded in view of Γ once we adopt BA.
Solution Take the minimal autoepistemic extensions!
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
102 / 104
Bootstrapping problem (Konolige (1988)) Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
The problem Take Γ = {BA ⊃ A}. There is an autoepistemic extension that contains A (and BA and A ∨ B and B(A ∨ B) and ¬B¬A, etc.). Note that A is grounded in view of Γ once we adopt BA.
Solution Take the minimal autoepistemic extensions! Note there is another autoepistemic extension Ξ0 of Γ that only contains (introspective) beliefs in classical tautologies and for all non-tautologies C it contains Bi ¬BC (where i ≥ 0). Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
102 / 104
Bootstrapping problem (Konolige (1988)) Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ})
The problem Take Γ = {BA ⊃ A}. There is an autoepistemic extension that contains A (and BA and A ∨ B and B(A ∨ B) and ¬B¬A, etc.). Note that A is grounded in view of Γ once we adopt BA.
Solution Take the minimal autoepistemic extensions! Note there is another autoepistemic extension Ξ0 of Γ that only contains (introspective) beliefs in classical tautologies and for all non-tautologies C it contains Bi ¬BC (where i ≥ 0). If we compare the subsets of Ξ and Ξ0 without occurrences of B, written Ξ0 and Ξ00 , (RUB, thenUGENT) Ξ00 ⊂ Ξ0Tutorial: . Christian Straßer Nonmonotonic Logic (Day 1) September 2, 2015 102 / 104
Bootstrapping problem continued
Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ}) Take Γ = {¬Bp ⊃ q, Bp ⊃ p}
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Bootstrapping problem continued
Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ}) Take Γ = {¬Bp ⊃ q, Bp ⊃ p} Two (minimal) autoepistemic extensions:
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Bootstrapping problem continued
Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ}) Take Γ = {¬Bp ⊃ q, Bp ⊃ p} Two (minimal) autoepistemic extensions: 1
including Bq, q and not including p
2
including Bp, p and not including q (and thus including ¬Bq)
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Bootstrapping problem continued
Recall: Ξ = Cn({Γ ∪ BΞ ∪ ¬BΞ}) Take Γ = {¬Bp ⊃ q, Bp ⊃ p} Two (minimal) autoepistemic extensions: 1
including Bq, q and not including p
2
including Bp, p and not including q (and thus including ¬Bq)
(see Konolige (1988))
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Topic 1
2 3
4 5
Defeasible Reasoning – Some Basic Concepts Some examples to warm up Some conceptual distinctions Nonmonotonic Logic in Context The Dynamics of Defeasible Reasoning Default Logic (in the tradition of Reiter) Warming up Defaults and Default Theories Inferring Alternatives and more examples Meta-Properties Introducing Priorities Summing up Autoepistemic Logic Bibliography Bibliography
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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Antonelli, G. A.: 1999, ‘A directly cautious theory of defeasible consequence for default logic via the notion of general extension’. Artificial Intelligence 109(1), 71–109. Gabbay, D. M.: 1985, ‘Theoretical foundations for non-monotonic reasoning in expert systems’. In: Logics and models of concurrent systems. New York, NY, USA: Springer-Verlag New York, Inc., pp. 439–457. Gordon, T., H. Prakken, and D. Walton: 2007, ‘The Carneades model of argument and burden of proof’. Artifical Intelligence 171(10), 875–896. Horty, J.: 2007, ‘Defaults with Priorities’. Journal of Philosophical Logic 36, 367–413. Horty, J. F.: 1994, ‘Some direct theories of nonmonotonic inheritance’. In: D. M. Gabbay, C. J. Hogger, and J. A. Robinson (eds.): Handbook of Logic in Artificial Intelligence and Logic Programming, Volume 3: Nonmonotonic Reasoning and Uncertain Reasoning. Oxford University Press, pp. 111–187. Horty, J. F.: 2012, Reasons as defaults. Oxford University Press. Konolige, K.: 1988, ‘On the relation between default and autoepistemic logic’. Artifical Intelligence 35(3), 343–382. Kraus, S., D. Lehman, and M. Magidor: 1990, ‘Nonmonotonic Reasoning, Preferential Models and Cumulative Logics’. Artifical Intelligence 44, 167–207. Łukaszewicz, W.: 1988, ‘Considerations on default logic: an alternative approach’. Computational intelligence 4(1), 1–16. Makinson, D. and P. Gärdenfors: 1991, ‘Relations between the logic of theory change and nonmonotonic logic’. In: The logic of theory change. Springer, pp. 183–205. Christian Straßer (RUB, UGENT)
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Moore, R. C.: 1984, ‘Possible-World Semantics for Autoepistemic Logic.’. In: Proceedings of the Workshop on non-monotonic reasoning. pp. 344–354. Moore, R. C.: 1985, ‘Semantical considerations on nonmonotonic logic’. Artifical Intelligence 25(1), 75–94. Pfeifer, N.: 2014, ‘Reasoning about uncertain conditionals’. Studia Logica pp. 1–18. Poole, D.: 1988, ‘A logical framework for default reasoning’. Artificial intelligence 36(1), 27–47. Poole, D.: 1991, ‘The effect of knowledge on belief: conditioning, specificity and the lottery paradox in default reasoning’. Artifical Intelligence 49(1-3), 281–307. Reiter, R.: 1980, ‘A Logic for Default Reasoning’. Artifical Intelligence 1–2(13). Stalnaker, R.: 1993, ‘A note on non-monotonic modal logic’. Artificial Intelligence 64(2), 183–196. Stalnaker, R.: 1994, ‘What is a nonmonotonic consequence relation?’. Fundamenta Informaticae 21(1), 7–21. Stenning, K. and M. Van Lambalgen: 2008, Human reasoning and cognitive science. MIT Press. Toulmin, S. E.: 1958, The Uses of Argument. Cambridge University Press.
Christian Straßer (RUB, UGENT)
Tutorial: Nonmonotonic Logic (Day 1)
September 2, 2015
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