SS10045: Proof Theory in Philosophy

Lecture 03: Natural Deduction I Dr Norbert Gratzl∗

Dr Ole Thomassen Hjortland† May 9, 2012

∗ Munich Center for Mathematical Philosophy (MCMP), Ludwig-Maximilians-Universität (LMU), 80539 München, Germany, [email protected]. † Munich Center for Mathematical Philosophy (MCMP), Ludwig-Maximilians-Universität (LMU), 80539 München, Germany, [email protected].

1

1 1.1

Natural Deduction Rules in Prawitz style and Lemmon style

The essential idea of natural deduction (what is supposed to make it ‘natural’) is that one works from assumptions. A proof in natural deduction is an array (presented either linearly, or in the form of a tree—a proof-tree) of wffs together with the assumptions on which those wffs depend. A sequent as defined above encodes a wff together with the assumptions on which it depends. We can present the rules pictorially (in the left-hand column below, following Gentzen and Prawitz, as parts of a proof-tree), or as a vertical sequence (in the right-hand column, following Lemmon, leading to a linear proof) as follows: Definition 1.1. We define the classical propositional natural deduction system NC as follows:

∧I: s1 , . . . , s m A B ∧ I A ∧ B

t1 , . . . , t n s1 , . . . , s m , t1 . . . , t n

(s) ... (t) ... (u)

A

...

B

...

A ∧ B

s,t ∧I

∧E: s1 , . . . , s m

(s) ... (t)

s1 , . . . , s m A ∧ B ∧ E A

A ∧ B ∧ E B

A ∧ B

...

A

s ∧E

A ∧ B

...

B

s ∧E

OR s1 , . . . , s m

(s) ... (t)

s1 , . . . , s m

∨I: s1 , . . . , s m s1 , . . . , s m A ∨I A∨B

B ∨I A∨B

(s) ... (t)

A

...

A∨B

s ∨I

B

...

A∨B

s ∨I

OR s1 , . . . , s m s1 , . . . , s m

2

(s) ... (t)

∨E: s1 , . . . , s m h

( A) ( B) A∨B C C ∨E C

t1 , . . . , t n t u1 , . . . , u o s1 . . . , (t1 , . . . tn −h),(u1 , . . . uo −t)

(g) ... (h) ... (s) ... (t) ... (u) ... (v)

A∨B

...

A

Ass

C

...

B

Ass

C

...

C

g,h,s,t,u ∨E

⊃I: ( A) B ⊃I A⊃B

s s1 , . . . , s m s1 , . . . , s m −s

(s) ... (t) ... (u)

A

Ass

B

...

A⊃B

s,t ⊃I

⊃E: s1 , . . . , s m A⊃B B

A

⊃E

t1 , . . . , t n s1 , . . . , t n

(s) ... (t) ... (u)

A⊃B

...

A

...

B

s,t ⊃E

¬I: ( A) ⊥ ¬I ¬A

s s1 , . . . , s m s1 , . . . , s m −s

3

(s) ... (t) ... (u)

A

Ass

⊥

...

¬A

s,t ¬I

¬E: s1 , . . . , s m

¬A A ¬E ⊥

t1 , . . . , t n s1 , . . . , t n

(s) ... (t) ... (u)

¬A

...

A

...

⊥

s,t ¬E

DN:

¬¬ A DN A

s1 , . . . , s m s1 , . . . , s m

(s) ... (u)

¬¬ A

...

A

s DN

Remark. The wff A in the rules ⊃I and ¬I, and the wffs A and B in the rule ∨E, which are removed from the assumptions in drawing the conclusion, are said to be discharged by the application of the rule. If there is a natural deduction proof of A from (undischarged) assumptions Σ in the system C (classical logic), we write Σ ` NC A. Proposition 1.1. NC is sound and complete. Definition 1.2. Intuitionistic propositional calculus NI results from NC as follows: (1) Drop DN (double negation elimination); and

⊥ (2) add ex false quodlibet (EFQ): A

1.2

Heuristics

Below is some strategic advice on how to derive in a natural deduction system. Sometimes it is helpful to work both from the top down, and from the bottom up. In the latter case, think about how the target formula can be derived (which rules to apply), and what you would require in order to do it. • If the sequent to be proved has the form Σ ` A, then start the proof by assuming, one on each line, each member of Σ • If at any point in the proof, one of the wffs to be used, e.g., one of the assumptions, has the form B ∨ C, and one’s aim is to derive D, then first assume B and derive D, then assume C and derive D, and finish by applying ∨E • If at some point one is trying to derive a wff of the form B ⊃ C, assume B, derive C and apply ⊃I 4

• If at some point one is trying to derive a wff of the form ¬ B, assume B, derive absurdity ⊥, and apply ¬I • If at some point one is trying to derive a wff of the form B ∧ C, first derive B, then derive C and apply ∧I • If at some point one cannot see how to derive some wff A (e.g., B ∨ C, where one cannot derive B or derive C from the given assumptions), try the ‘¬I + DN’-strategy: assume ¬ A, deduce a contradiction (⊥), use ¬I to infer ¬¬ A (discharging the assumption ¬ A), then use DN to infer A.

1.3

Examples

1 2 3 1,3 1,2,3 1,2 1

(1) (2) (3) (4) (5) (6) (7)

p⊃q ¬q p q ⊥ ¬p ¬q ⊃ ¬ p

Ass (Premise) Ass (for ⊃I) Ass (for ¬I) 1,3 ⊃E 2,4 ¬E 3,5 ¬I 2,6 ⊃I

So, p ⊃ q ` NC ¬q ⊃ ¬ p 1 2 2 2 2 6 6 6 6 6 6 1

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)

p ∨ (q ∧ r ) p p∨q p∨r ( p ∨ q) ∧ ( p ∨ r ) q ∧ r q p∨q r p∨r ( p ∨ q) ∧ ( p ∨ r ) ( p ∨ q) ∧ ( p ∨ r )

Ass (Premise) Ass (for ∨E) 2 ∨I 2 ∨I 3,4 ∧ I Ass (for ∨E) 6 ∧ E 7 ∨I 6 ∧ E 9 ∨I 8,10 ∧ I 1,2,5,6,11 ∨E

So p ∨ (q ∧ r ) ` NC ( p ∨ q) ∧ ( p ∨ r )

1.4

Exercises

Exercise 1: Give natural deduction proofs to show the following.: 1. ` NC ( p ⊃ ¬ p) ⊃ ¬ p 2. ¬ p ` NC p ⊃ q

5

3. ¬ p ⊃ p ` NC p 4. p ⊃ r, q ⊃ r ` NC ( p ∨ q) ⊃ r 5. ` NC p ∨ ¬ p Try to give the derivations in both Lemmon style and Prawtiz tree style. Exercise 2: The classical biconditional ↔ is defined as A ↔ B := ( A → B) ∧ ( B → A). Try to devise natural deduction rules for the biconditional (introduction rules and elimination rules). It is not required that you give a (formal) soundness and completeness proof. Exercise 3: Intuitionistic negation can be formalized in a number of ways. Instead of the rules in NI we can give rules without using ⊥. Show that the two following set of inference rules for intuitionistic negation are equivalent:

[ A] .. .. (1) ¬A ¬A

A

[ A] [ A] .. .. .. .. (2) B ¬B ¬A

6

¬A B

A

¬A B