31 October 2015 T. Scott Dixon Ashoka University

Supplementary Proofs to “What Is the Well-Foundedness of Grounding?”1 What follows are proofs of several claims made in Dixon forthcoming. px ← Γq abbreviates px is fully grounded by Γq, and px | ← Γq abbreviates px is partially grounded by Γq. pΓ  ∆q abbreviates pΓ are among ∆q. ‘F’ plurally denotes the (non-empty) domain of facts. Preliminary Proofs The first proofs support claims made in section 1.2 Partial Transitivity. For any x, y, and z, if x |← y and y |← z, then x |← z. Proof. Consider arbitrary facts a, b, and c, and suppose that a |← b and b |← c. By the definition of partial grounding, a ← b, Γ for some Γ and b ← c, ∆ for some ∆. By full transitivity, a ← Γ, c, ∆. By the definition of partial grounding, a |← c. Partial Irreflexivity. For any x, ∼ x |← x. Proof. Consider an arbitrary fact a, and suppose for reductio that a |← a. By partial asymmetry, ∼ a |← a. Contradiction. So ∼ a |← a. Full Asymmetry. For any x annd y, if x ← y, then ∼ y ← x. Proof. Consider arbitrary facts a and b, and suppose that a ← b. By the definition of partial grounding, a |← b. By partial asymmetry, ∼ b |← a. So there are no Γ such that ∼ b ← Γ and a  Γ. Since a  a, ∼ b ← a. Full Irreflexivity. For any x, ∼ x ← x. Proof. Consider an arbitrary fact a, and suppose for reductio that a ← a. Since a  a, the definition of partial grounding guarantees that a |← a. By partial asymmetry, ∼ a |← a. Contradiction. So ∼ a ← a.

1 2

Many thanks to Aldo Antonelli and Cody Gilmore for their careful readings of these proofs. All section and footnote references are to Dixon forthcoming.

1

Proof 1 The first substantial proof supports a claim made at the end of section 1. Proposition 1. For any grounding chain Γ, Γ are a non-terminating grounding chain iff for any x  Γ, there is a y  Γ such that x |← y. Requisite Definition: (Downwardly) Terminating Grounding Chains. Γ are a (downwardly) terminating grounding chain =df (i) Γ are a grounding chain and (ii) there is a y  Γ such that, for any x  Γ, if x 6= y then x |← y. So a non-terminating grounding chain Γ are a grounding chain such that there is no y  Γ such that, for any x  Γ, if x 6= y then x |← y. Proof. Consider an arbitrary grounding chain B (‘beta’). (⇒) Suppose first that B are a non-terminating grounding chain. Then there is no y  B such that for any x  B, if x 6= y then x |← y. That is, for any y  B, there is an x  B such that x 6= y and ∼ x |← y. Now suppose for reductio that it is not the case that for any x  B, there is a y  B such that x |← y. That is, there is an x  B such that for any y  B, ∼ x |← y. Let b be some such fact. Then b  B and for any y  B, ∼ b |← y. By the last result before the reductio assumption, there is an x  B such that x 6= b and ∼ x |← b. Let c be some such fact. Then c 6= b and ∼ c |← b. Since B are a grounding chain, partial grounding is connected over B, and so for any x  B and any y  B, either x |← y, y |← x, or x = y. In particular, either c |← b, b |← c, or c = b. It has already been established ∼ c |← b and that c 6= b. So b |← c. But it has also been established that for any y  B, ∼ b |← y, and so ∼ b |← c. Contradiction. So for any x  B, there is a y  B such that x |← y. (⇐) Now suppose that for any x  B, there is a y  B such that x |← y. Suppose for reductio that B are a terminating grounding chain. Then there is a y  B such that for any x  B, if x 6= y then x |← y. Let b be some such fact. Then b  B and for any x  B, if x 6= b then x |← b. By the supposition just before the reductio assumption, there is a y  B such that b |← y. Let c be some such fact. Then c  B and b |← c. Since b |← c, the irreflexivity of grounding guarantees that b 6= c. But since also c  B, c |← b. This contradicts partial asymmetry. So B are a non-terminating grounding chain. Proof 2 The next proof supports a claim made in section 4 (see fn. 19). Proposition 2. If every grounding chain terminates, then every non-fundamental fact is fully grounded by some fundamental facts. ((S) implies (FS).) 2

Requisite Definitions and Principles: Plural Complements. Γ − ∆ =df the E such that x  E iff (i) x  Γ and (ii) x  ∆. Plural Unions.

[

Γi =df the ∆ such that x  ∆ iff x  Γi for some Γi where i ≤ κ.

i≤κ

Generalized Full Transitivity. For any cardinal number κ, any x and Γ, and any yi ’s and ∆i ’s for all i ≤ κ, if x is fully grounded by the yi ’s for all i ≤ κ, together [with Γ, and yi is fully grounded by ∆i for each i ≤ κ, then x is fully grounded by Γ, ∆i .3 i≤κ

Principle of Dependent Choices. If R is a binary relation on Γ, and if for any x  Γ, there is a y  Γ such that Rxy, then there is a sequence x0 , x1 , . . . , xn , . . .  Γ such that Rxn xn+1 for every n ∈ N.4 In particular, For any Γ  F, if for any x  Γ, there is a y  Γ such that x |← y, then there are x0 , x1 , . . . , xn , . . .  Γ such that xn |← xn+1 for every n ∈ N.5 Transitivity of Inclusion. For any Γ, ∆, and E (‘epsilon’), if Γ  ∆ and ∆  E, then Γ  E.

Proof. Suppose that every grounding chain terminates, and consider an arbitrary non-fundamental fact a. Suppose for reductio that a is not fully grounded by any fundamental facts, and consider the facts Z such that for any x, x  Z iff (i) a |← x, (ii) x is not fundamental, and (iii) x is not fully grounded by any fundamental facts. I now show that Z exist. Since a is not fundamental, a is partially, and so (by the definition of partial grounding) fully grounded by some facts. Let Ha be some such facts. Then a is fully grounded by Ha . And since a is not fully grounded by any fundamental facts, there are Θa  Ha such that for any x  Θa , x is not fundamental. Suppose for reductio that for any y  Θa , y is fully grounded by some fundamental facts, and suppose that there are κ facts among Θa . Now let the bi ’s for all i ≤ κ be all and only the facts among Θa , and let Ibi be some fundamental facts that fully ground bi for each i ≤ κ. (The last reductio assumption guarantees that Ibi exist for each i ≤ κ.) This means that a is fully grounded by the bi ’s for all i ≤ κ, together with Ha − Θa . And since bi is fully grounded by Ibi for each i ≤ κ, the [ transitivity of full grounding guarantees that a is fully grounded by Ha − Θa , Ibi . However, by the characterizations i≤κ 3

This principle is a generalization of full transitivity (see section 2). Adapted from Jech 2002: 50. 5 Partial grounding is a relation on F because grounding is a relation on S, where for any x, x ∈ S iff x  F. 4

3

of each of the Ibi ’s for all i ≤ κ, Ha , and Θa , every fact among Ha − Θa , is fundamental. And, importantly, since a is fully grounded by Ha −Θa ,

[

Ibi

i≤κ [ Ibi , i≤κ

there is an x  Ha − Θa ,

[

Ibi such that a |← x, and so Ha − Θa ,

[

Ibi exist.

i≤κ

i≤κ

But then a is fully grounded by some fundamental facts, contradicting the initial reductio assumption. So there is a y  Θa such that y is not fully grounded by any fundamental facts. Let b be some such fact. Then b  Θa and b is not fully grounded by any fundamental facts. Moreover, since b  Θa , a |← b and b is not fundamental. By our characterization of Z, b  Z. As a result, Z exist. I now show that for any x  Z, there is a y  Z such that x |← y. Consider an arbitrary c  Z. By the characterizaton of Z, a |← c, c is not fundamental, and c is not fully grounded by any fundamental facts. Since c is not fundamental, c is partially, and so (by the definition of partial grounding) fully grounded by some facts. Let Hc be some such facts. Then c is fully grounded by Hc . And since c is not fully grounded by any fundamental facts, there are Θc  Hc such that for any x  Θc , x is not fundamental. Suppose for reductio that for any y ⊆ Θc , y is fully grounded by some fundamental facts, and suppose that there are κ facts among Θc . Now let the di ’s for all i ≤ κ be all and only the facts among Θc , and let Idi be some fundamental facts that fully ground di for each i ≤ κ. (The last reductio assumption guarantees that Idi exist for each i ≤ κ.) This means that c is fully grounded by the di ’s for all i ≤ κ, together with Hc − Θc . And since of full grounding di is fully grounded by Idi for each i ≤ κ, the transitivity [ guarantees that, c is fully grounded by Hc − Θc , Idi . However, by the chari≤κ

acterizations [ of each of the Idi ’s for all i ≤ κ, Hc , and Θc , every fact among Hc − Θc , Idi is fundamental. And, importantly, since c is fully grounded i≤κ

by Hc − Θc ,

[

Idi , there is an x  Hc − Θc ,

[

Idi such that c |← x, and so

i≤κ

i≤κ

Hc − Θc ,

[

Idi exist. But then c is fully grounded by some fundamental facts,

i≤κ

contradicting an earlier result obtained before the last reductio assumption was made. So there is a y  Θc such that y is not fully grounded by any fundamental facts. Let d be some such fact. Then d  Θc and d is not fully grounded by any fundamental facts. Moreover, since d  Θc , c |← d and d is not fundamental. And since a |← c and c |← d, the transitivity of partial grounding guarantees that a |← d. By our characterization of Z, d  Z. So there is a y  Z such that c |← y. And since c is arbitrary, for any x  Z, there is a y  Z such that x |← y. I now show that there is a non-terminating grounding chain among Z. Recall that Z exist. And since Z includes only facts, Z  F. By the above instance of the principle 4

of dependent choices, there are x0 , x1 , . . . , xn , . . .  Z such that xn |← xn+1 for every n ∈ N. Let a0 , a1 , . . . , an , . . . be some such facts. Then a0 , a1 , . . . , an , . . .  Z and an |← an+1 for every n ∈ N. Next suppose that for any x, x  ZC iff x  a0 , a1 , . . . , an , . . .. I now show that ZC are a grounding chain. To accomplish this, I first show that ZC are a grounding structure. Since Z  F and ZC  Z, the transitivity of inclusion guarantees that ZC  F. Further, since 0 ∈ N, a0  ZC . By the above characterization of Z, then, a0 |← a1 . And since 1 ∈ N, a1  ZC . So for some x  ZC , there is a y  ZC such that x |← y. Thus ZC are a grounding structure. I now show that partial grounding is connected over ZC . Consider arbitrary ai , aj  ZC , where i, j ∈ N. Either (i) ai = aj or (ii) ai 6= aj . (i) Suppose first that ai = aj . Then either ai |← aj , aj |← ai , or ai = aj . (ii) Now suppose that ai 6= aj . Then i 6= j, and so either (a) i < j or (b) j < i. (a) Suppose that i < j. I show by induction that for every n ∈ N, if i < n, then ai |← an . Basis. By the above characterization of ZC , ai |← ai+1 . Inductive Step. Suppose that ai |← am , where m ∈ N. By the above characterization of ZC , am | ← am+1 . By the transitivity of partial grounding, ai |← am+1 . So ai |← aj . Hence either ai |← aj , aj |← ai , or ai = aj . (b) Now suppose that j < i. I show by induction that for all n ∈ N, if j < n, then aj |← an . Basis. By the above characterization of ZC , aj |← aj+1 . Inductive Step. Suppose that aj |← am , where m ∈ N. By the above characterization of ZC , am | ← am+1 . By the transitivity of partial grounding, aj |← am+1 . So aj |← ai . Hence either ai |← aj , aj |← ai , or ai = aj . Hence, whether i < j or j < i, either ai |← aj , aj |← ai , or ai = aj . And, whether ai = aj or ai 6= aj , either ai |← aj , aj |← ai , or ai = aj . Since ai and aj are arbitrary, for any x, y  ZC , either x |← y, y |← x, or x = y. By the definition of connectedness, partial grounding is connected over ZC . It now follows that ZC are a grounding chain. I now show that ZC do not terminate. Suppose for reductio that ZC terminate. Then there is a y  ZC such that for any x  ZC , if x 6= y, then x |← y. Let e be some such fact. Then for any x  ZC , if x 6= e, then x |← e. Now suppose for reductio that there is a y  ZC such that e |← y. Let f be some such fact. Then f  ZC and e |← f . By the irreflexivity of partial grounding, e 6= f . And since f  ZC , the result just before the last reductio assumption guarantees that f |← e. By the transitivity of partial grounding, e |← e, contradicting the irreflexivity of partial grounding. 5

So there is no y  ZC such that e | ← y. But now consider an arbitrary ai  ZC , where i ∈ N. By the above characterization of ZC , ai |← ai+1 . Since i ∈ N, i + 1 ∈ N. By the characterization of ZC , ai+1  ZC . So there is a y  ZC such that ai |← y. Since ai is arbitrary, for any x  ZC , there is a y  ZC such that x |← y. So there is a y  ZC such that e |← y. This contradicts the earlier claim that there is no y  ZC such that e |← y. So ZC are a non-terminating grounding chain. But since ZC are a grounding chain, the initial assumption guarantees that ZC are a terminating grounding chain. Contradiction. Hence a is fully grounded by some fundamental facts. And since a is arbitrary, every nonfundamental fact is fully grounded by some fundamental facts. Proof 3 The next proof supports a claim made in section 5 (see fn. 28).6 Proposition 3. Every grounding chain terminates iff for any non-empty set of facts S, S has a minimal ground. ((S) and (WF) are equivalent.) Requisite Definition: Minimal Grounds. A set S has a minimal ground =df there is a y ∈ S such that, for any x ∈ S, ∼ y |← x.

Proof. Blah. (⇒) Suppose that every grounding chain terminates. Suppose for reductio that some non-empty set of facts S has no minimal ground. Then there is no y ∈ S such that for any x ∈ S, ∼ y |← x. Suppose that for any x, x  A (‘alpha’) iff x ∈ S. Then there is no y  A such that for any x  A, ∼ y |← x. That is, for any y  A, there is an x  A such that y |← x. Since S is non-empty, A exist. And since A includes only facts, A  F. By the above instance of the principle of dependent choices, there are x0 , x1 , . . . , xn , . . .  A such that xn |← xn+1 for every n ∈ N. Let a0 , a1 , . . . , an , . . . be some such facts. Then a0 , a1 , . . . , an , . . .  A and an |← an+1 for every n ∈ N. Next suppose that for any x, x  B (‘beta’) iff x  a0 , a1 , . . . , an , . . .. I now show that B are a grounding chain. To accomplish this, I first show that B are a grounding structure. Since S is a nonempty set of facts, A  F. And since B  A, the transitivity of inclusion guarantees that B  F. Further, since 0 ∈ N, a0  B. By the above characterization of B, then, a0 |← a1 . And since 1 ∈ N, a1  B. So for some x  B, there is a y  B such that x |← y. Thus B are a grounding structure. I now show that partial grounding is connected over B. Consider arbitrary ai , aj  B, where i, j ∈ N. Either (i) ai = aj or (ii) ai 6= aj . 6

The proof closely follows a proof found in many set theory texts. See, for example, Jech 2002: 50-51.

6

(i) Suppose first that ai = aj . Then either ai |← aj , aj |← ai , or ai = aj . (ii) Now suppose that ai 6= aj . Then i 6= j, and so either (a) i < j or (b) j < i. (a) Suppose that i < j. I show by induction that for every n ∈ N, if i < n, then ai |← an . Basis. By the above characterization of B, ai |← ai+1 . Inductive Step. Suppose that ai |← am , where m ∈ N. By the above characterization of B, am | ← am+1 . By the transitivity of partial grounding, ai |← am+1 . So ai |← aj . Hence either ai |← aj , aj |← ai , or ai = aj . (b) Now suppose that j < i. I show by induction that for all n ∈ N, if j < n, then aj |← an . Basis. By the above characterization of B, aj |← aj+1 . Inductive Step. Suppose that aj |← am , where m ∈ N. By the above characterization of B, am | ← am+1 . By the transitivity of partial grounding, aj |← am+1 . So aj |← ai . Hence either ai |← aj , aj |← ai , or ai = aj . Thus, whether i < j or j < i, either ai |← aj , aj |← ai , or ai = aj . And, whether ai = aj or ai 6= aj , either ai |← aj , aj |← ai , or ai = aj . Since ai and aj are arbitrary, for any x  B and any y  B, either x |← y, y |← x, or x = y. By the definition of connectedness, partial grounding is connected over B. It now follows that B are a grounding chain. I now show that B do not terminate. Consider an arbitrary ai  B, where i ∈ N. By the above characterization of B, ai |← ai+1 . Since i ∈ N, i + 1 ∈ N. By the characterization of B, ai+1  B. So there is a y  B such that ai |← y. And since ai is arbitrary, for any x  B, there is a y  B such that x |← y. Since B is a grounding chain, proposition 1 guarantees that B do not terminate. But the initial assumption guarantees that B terminate. Contradiction. So every non-empty set of facts has a minimal ground. (⇐) Suppose that for any non-empty set S, S has a minimal ground. Then for any non-empty set S, there is a y ∈ S such that for any x ∈ S, ∼ y |← x. That is, for any Γ, there is a y  Γ such that for any x  Γ, ∼ y |← x. Now consider an arbitrary grounding chain B (‘beta’). It follows that there is a y  B such that for any x  B, ∼ y |← x.7 Let m be some such fact. Then for any x  B, ∼ m |← x. Since B are a grounding chain, partial grounding is connected over B, and so for any x  B, either m |← x, x |← m, or m = x. Since no x  B partially grounds m, for any x  B, either x |← m, or m = x. That is, for any x  B, if x 6= m, then x |← m. So there is a y  B such that, for any x  B, if x 6= y, then x |← y. Hence B are a terminating grounding chain. And since B are arbitrary, every grounding chain terminates.

7

If there are pluralities larger than sets, then this proof is valid only for grounding chains that are set-sized.

7

Proof 4 The next proof supports a claim made in fn. 30. Proposition 4. Every maximal grounding chain terminates iff every grounding chain has a lower bound. ((M) and (L) are equivalent.) Requisite Definitions and Principle: (Downwardly) ∆-Maximal Grounding Chains. Γ are a (downwardly) ∆-maximal grounding chain =df (i) Γ  ∆, (ii) Γ are a grounding chain, and (iii) there is no y  ∆ that partially grounds every x  Γ. (Downwardly) Maximal Grounding Chains. Γ are a (downwardly) maximal grounding chain =df Γ are a (downwardly) F-maximal grounding chain. Hausdorff Maximal Principle. For any Γ, if Γ are partially ordered, then for any chain ∆  Γ, there are E such that (i) E are a Γ-maximal chain, (ii) ∆  E, and (iii) E  Γ.8 Proof. (⇒) Suppose that every maximal grounding chain terminates, and consider an arbitrary grounding chain A (‘alpha’). Since A are a grounding chain, the Hausdorff maximal principle guarantees that there is a maximal grounding chain M (‘mu’) such that A  M. By the initial assumption, every maximal grounding chain terminates. So there is a y  M such that for any x  M, if x 6= y, then x |← y. Let b be some such fact. Then b  M and for any x  M, if x 6= b, then x |← b. Now consider an arbitrary a  A. Since A  M, the transitivity of inclusion guarantees that a  M. Since for any x  M, if x 6= b, then x |← b, if a 6= b, then a |← b. In other words, either a = b or a |← b. Since a is arbitrary, for any x  A, either x = b or x |← b. So b is a lower bound of A. And since A are arbitrary, every grounding chain has a lower bound. (⇐) Now suppose that every grounding chain has a lower bound, and consider an arbitrary maximal grounding chain M (‘mu’). Since M is a grounding chain, the initial assumption guarantees that M has a lower bound. Let b be some such fact. Then for any x  M, either x = b or x |← b. I now show that b  M. Suppose for reductio that b  M, and consider an arbitrary fact a  M. Since b  M, the indiscernibility of identicals guarantees that a 6= b, and so, by the last result before the reductio assumption, a |← b. Since a is arbitrary, for any x  M, x |← b. So there is a y such that for any x  M, x |← y. Thus M is not maximal, contradicting one of the initial assumptions. It now follows that b  M. As a result, there is a y  M such that for any x  M, either x = y or x |← y. In other words, there is a y  M such that for any x  M, if x 6= y, then x |← y. So M is a terminating grounding chain. And since M is arbitrary, every maximal grounding chain terminates. 8

A proof of the Hausdorff maximal principle requires the axiom of choice or an equivalent principle, such as the well-ordering theorem. In fact, the Hausdorff maximal principle is equivalent to the axiom of choice. For a proof of the Hausdorff maximal principle using the well-ordering theorem, see Hausdorff 1914: 140-41 or 1962: 197-98. For a proof using the axiom of choice itself, see Frink 1952. Devlin (1979: 56-63) shows that the Hausdorff maximal principle is equivalent to the axiom of choice as well as several variants of it.

8

Proof 5 The next proof supports a claim made in section 7 (see fn. 36). Proposition 5. If every weakly maximal grounding chain terminates, then every nonfundamental fact is fully grounded by some fundamental facts. ((WM) implies (FS).) Proof. Suppose that every weakly maximal grounding chain terminates, and consider an arbitrary non-fundamental fact a. Suppose for reductio that a is not fully grounded by any fundamental facts, and consider the facts Z (‘zeta’) such that for any x, x  Z iff (i) a |← x, (ii) x is not fundamental, and (iii) x is not fully grounded by any fundamental facts. I first establish that Z exist. Since a is not fundamental, a is partially, and so (by the definition of partial grounding) fully grounded by some facts. Let Ha (‘eta sub a’) be some such facts. Then a is fully grounded by Ha . And since a is not fully grounded by any fundamental facts, there are Θa (‘theta sub a’)  Ha such that for any x  Θa , x is not fundamental. Suppose for reductio that for any y  Θa , y is fully grounded by some fundamental facts, and suppose that there are κ facts among Θa . Now let the bi ’s for all i ≤ κ be all and only the facts among Θa , and let Ibi (‘iota sub (b sub i)’) be some fundamental facts that fully ground bi for each i ≤ κ. (The most recent reductio assumption guarantees that Ibi exist for each i ≤ κ.) This means that a is fully grounded by the bi ’s for all i ≤ κ, together with Ha − Θa . And since bi is fully grounded by Ibi for each i ≤ κ, the transitiv[ ity of full grounding guarantees that a is fully grounded by (Ha − Θa ), Ibi . i≤κ

However, by the characterizations [ of each of the Ibi ’s for all i ≤ κ, Ha , and Θa , every fact among (Ha − Θa ), Ibi is fundamental. And importantly, since i≤κ

Θa exist, some y[  Θa exists, and so Ibi exist for[some i ≤ κ. And since Ibi  (Ha − Θa ), Ibi for every i ≤ κ, (Ha − Θa ), Ibi exist. But then a is i≤κ

i≤κ

fully grounded by some fundamental facts, contradicting the initial reductio assumption. So there is a y  Θa such that y is not fully grounded by any fundamental facts. Let b be some such fact. Then b  Θa and b is not fully grounded by any fundamental facts. Moreover, since b  Θa , a |← b and b is not fundamental. By our characterization of Z, b  Z. As a result, Z exist. I now show that there is a Z-maximal grounding chain among Z. Since Z exist, there is an x  Z. Let c be some such fact. By the characterizaton of Z, c is not fundamental, and so c is partially, and so (by the definition of partial grounding) fully grounded by some facts. Let Hc (‘eta sub c’) be some such facts. Then c is fully grounded by Hc . The characterization of Z also guarantees that c is not fully grounded by any fundamental facts, and so there are Θc (‘theta sub c’)  Hc such that for any x  Θc , x is not fundamental. 9

Suppose for reductio that for any y  Θc , y is fully grounded by some fundamental facts, and suppose that there are κ facts among Θc . Now let the di ’s for all i ≤ κ be all and only the facts among Θc , and let Idi (‘iota sub (d sub i)’) be some fundamental facts that fully ground di for each i ≤ κ. (The most recent reductio assumption guarantees that Idi exist for each i ≤ κ.) This means that c is fully grounded by the di ’s for all i ≤ κ, together with Hc − Θc . And since di is fully grounded by Idi for each i ≤ κ, the transitiv[ ity of full grounding guarantees that, c is fully grounded by Hc − Θc , Idi . i≤κ

However, by the characterizations [ of each of the Idi ’s for all i ≤ κ, Hc , and Θc , every fact among Hc − Θc , Idi is fundamental. And importantly, since i≤κ

Θc exist, some y [  Θc exists, and so Idi exist for some [ i ≤ κ. And since Idi  (Hc − Θc ), Idi for every i ≤ κ, (Hc − Θc ), Idi exist. But then c i≤κ

i≤κ

is fully grounded by some fundamental facts, contradicting an earlier result obtained before the last reductio assumption was made. So there is a y  Θc such that y is not fully grounded by any fundamental facts. Let d be some such fact. Then d  Θc and d is not fully grounded by any fundamental facts. Moreover, since d  Θc , c |← d and d is not fundamental. And since a |← c and c |← d, the transitivity of partial grounding guarantees that a |← d. By our characterization of Z, d  Z. Since b |← c, b, c is a grounding structure. And since b and c are the only facts among b, c, for any x  b, c and any y  b, c, either x |← y, y |← x, or x = y. So b, c are a grounding chain. Since b, c  Z, and Z are partially ordered by grounding, the Hausdorff maximal principle guarantees that there is a Z-maximal grounding chain M (‘mu’) such that b, c  M  Z. I now show that M are weakly maximal. Suppose for reductio that M are not weakly maximal. Then there are ∆ such that every x among some final segment MF (‘mu sub F’)  M is fully grounded by ∆. Let Λ (‘lambda’) be some such facts. Then every x  MF is fully grounded by Λ. Now since M exist, M have a final segment, and so MF exist. That is, there is an x  MF . Let e be some such fact. Then e  MF and e ← Λ. For convenience, let Λ = He (‘eta sub e’). I now show that some x  He is not fully grounded by any fundamental facts. Suppose there is no x  He that is not fundamental and is not fully grounded by any fundamental facts. I now show that it follows that e is fully grounded by some fundamental facts. Suppose e is not fully grounded by some fundamental facts. Then there are Θe (‘theta sub e’)  He such that for any x  Θe , x is not fundamental. Suppose for reductio that for any y  Θe , y is fully grounded by some fundamental facts, and suppose that there are κ facts among Θe . Now let the fi ’s for all i ≤ κ be all and only the facts among Θe , and let Ifi (‘iota sub (f sub i)’) be some fundamental facts that fully ground 10

fi for each i ≤ κ. (The most recent reductio assumption guarantees that Ifi exist for each i ≤ κ.) This means that e is fully grounded by the fi ’s for all i ≤ κ, together with He − Θe . And since fi is fully grounded by Ifi for each i ≤ κ, the transitivity of full grounding guarantees that [ e is fully grounded by He − Θe , Ifi . However, by the i≤κ

characterizations of each of the[ Ifi ’s for all i ≤ κ, He , and Θe , every fact among He −Θe , Ifi is fundamental. And i≤κ

importantly, since Θe exist, some y  Θe exists, and [ so Ifi exist for some i ≤ κ. And since Ifi  (He −Θe ), Ifi i≤κ

for every i ≤ κ, (He −Θe ),

[

Ifi exist. But then e is fully

i≤κ

grounded by some fundamental facts, contradicting the most recent reductio assumption. So there is a x  Θe such that x is not fully grounded by any fundamental facts. Let f be some such fact. Then f  Θe and f is not fully grounded by any fundamental facts. Morover, since f  Θe , f is not fundamental. And since Θe  He , there is an x  He that is not fundamental and is not fully grounded by any fundamental facts. But this contradicts an earlier reductio assumption. So e is fully grounded by some fundamental facts. But, since MF  M  Z, e  Z. By the characterization of Z, e is not fully grounded by any fundamental facts. Contradiction. So there is an x  He that is not fundamental and is not fully grounded by any fundamental facts. Let g be some such fact. Then g  He , g is not fundamental, and g is not fully grounded by any fundamental facts. Since MF  M  Z, a is partially grounded by every x  MF . And since every x  MF is fully grounded by Λ = He , every x  MF is partially grounded by g. So the transitivity of partial grounding guarantees that a |← g. By the characterization of Z, g  Z. So some y  Z partially grounds every x among some final segment of M. As a result, M are not Z-maximal, contradicting an earlier result. So M are weakly maximal. That is, there are no ∆ that fully ground every x among any final segment of M. I now show that M do not terminate. Suppose for reductio that M terminate. Then there is a y  M such that, for every x  M such that x 6= y, x |← y. Let m be some such fact. Then for every x  M such that x 6= m, x |← m. But since m  M and M  Z, m  Z. By the characterization of Z, m is not fundamental. This means that, for some y, m |← y. Let n be some such fact. Then m |← n. By the definition of partial grounding, there are ∆ such that m ← ∆. Since m is a final segment of M, there is a final segment of M that is fully grounded by some facts. So M is not weakly maximal, contradicting the earlier result. 11

So M do not terminate. But since M is weakly maximal, the initial assumption guarantees that M terminate. Contradiction. Hence a is fully grounded by some fundamental facts. And since a is arbitrary, every nonfundamental fact is fully grounded by some fundamental facts. Proof 6 The next proof supports a claim made in fn. 39. Proposition 6. Every non-fundamental fact is partially grounded by some fundamental fact iff every non-fundamental fact is partially grounded by a fact with which it does not share a partial ground. ((PF) and (R) are equivalent.)

Proof. (⇒) Suppose that every non-fundamental fact is partially grounded by some fundamental fact, and consider an arbitrary non-fundamental fact a. It follows that a is partially grounded by some fundamental fact. Let b be some such fact. Then b is fundamental and a |← b. Suppose for reductio that there is a z such that both b |← z and a |← z. Let c be some such fact. Then b |← c and a |← c. Since b |← c, b is not fundamental, contradicting our earlier result. So there is no z such that both b |← z and a |← z. Hence, every non-fundamental fact is partially grounded by a fact with which it does not share a partial ground. (⇐) Suppose that every non-fundamental fact is partially grounded by a fact with which it does not share a partial ground, and consider an arbitrary non-fundamental fact a. It follows that a is partially grounded by a fact with which it does not share a partial ground. Let b be some such fact. Then a |← b and there is no z such that both b |← z and a |← z. Suppose for reductio that a is not partially grounded by any fundamental fact. Since a |← b, b is not fundamental, and so b is partially grounded by some fact. Let c be some such fact. Then b |← c. By the transitivity of partial grounding, a |← c. So b |← c and a |← c. As a result, there is a z such that both b |← z and a |← z, contradicting our earlier result. So a is partially grounded by some fundamental fact. Hence, every non-fundamental fact is partially grounded by some fundamental fact. Proof 7 The next proof supports a claim made in section 8 (see fn. 40). Proposition 7. If every maximal grounding chain terminates, then every non-fundamental fact is partially grounded by some fundamental fact. ((M) implies (PF).)

Proof. Suppose that every maximal grounding chain terminates, and consider an arbitrary nonfundamental fact a. 12

Suppose for reductio that a is not partially grounded by any fundamental fact, and consider the facts Z (‘zeta’) such that for any x, x  Z iff a | ← x. Since a is not fundamental, Z exist. I now show that for any x  Z, there is a y  Z such that x |← y. Consider an arbitrary b  Z. Suppose for reductio that there is no y  Z such that b |← y. That is, for any y  Z, ∼ b |← y. Now either (i) b is fundamental or (ii) b is not fundamental. (i) Suppose first that b is fundamental. Then, since a |← b, a is partially grounded by a fundamental fact, contradicting the first reductio assumption. (ii) Suppose next that b is not fundamental. Then there is a y such that b |← y. Let c be some such fact. Then b |← c. And since a |← b, the transitivity of partial grounding guarantees that a |← c. By the above characterization of Z, b  Z. But by the last reductio assumption, c  Z. Contradiction. Either way, then, there is a contradiction. So there is a y  Z such that b |← y. And since b is arbitrary, for any x  Z, there is a y  Z such that x |← y. Next, I show that there is a maximal grounding chain among Z. Recall that b  Z. From the last result, it follows that there is a y  Z such that b |← y. Let c be some such fact. Then c  Z and b |← c. So there is a chain ZC  Z. And since ZC are a grounding chain, the Hausdorff maximal principle guarantees that there is a maximal grounding chain M (‘mu’) such that ZC  M. Since b  ZC , b  M. Suppose that for any x, x  N (‘nu’) iff x  M and b |← x. Since c exists and c  M, N exist. Further, since M is maximal, the characterization of N guarantees that N is maximal. I now show that N  Z. Consider an arbitrary d  N. By the characterization of N, b |← d. Since a |← b, the transitivity of partial grounding guarantees that a |← d. By the characterization of Z, d  Z. And since d is arbitrary, for any x  N, x  Z. So N  Z. Next, I show that for every x  N, there is a y  N such that x |← y. Consider an arbitrary d  N. Suppose for reductio that there is no y  N such that d |← y. That is, for every y  N, ∼ d |← x. Since N is a grounding chain, grounding is connected over N, and so for any y  N, either d = y or y |← d. In other words, for any y  N, if d 6= y, then y |← d. Since d  N and N  Z, d  Z. And since for every x  Z, there is a y  Z such that x |← y, there is a y  Z such that d |← y. Let e be some such fact. Then e  Z and d |← e. Next consider an arbitrary f  N. Either d = f or d 6= f . (i) Suppose first that d = f . Then f |← e. (ii) Suppose next that f 6= d, and recall that, for any y  N, if d 6= y, then y |← d. It follows that f |← d. By partial transitivity, f |← e. 13

Either way, then, f |← d. And since f is arbitrary, for every x  N, x |← e. So there is a y such that for every x  N, x |← y. This means that N are not maximal, contradicting the earlier result. So for every x  N, there is a y  N such that x | ← y. By proposition 1, N do not terminate. But, since N are maximal, the initial assumption guarantees that N terminate. Contradiction. Hence a is partially grounded by some fundamental fact. And since a is arbitrary, every nonfundamental fact is partially grounded by some fundamental fact.

References Devlin, K. 1979. The Joy of Sets, 2nd Ed. New York: Springer. Dixon, T. S. forthcoming. What is the well-foundedness of grounding? Mind. Frink, O. 1952. A proof of the maximal chain theorem. American Journal of Mathematics 74, 676–78. Hausdorff, F. 1914. Grundz¨ uge der Megnenlehre. Leipzig. Hausdorff, F. 1962. Set Theory. New York: Chelsea Publishing Company. Jech, Thomas 2002. Set Theory: The Third Millenium Edition. Berlin: Springer-Verlag, rpnt 2006. Originally published as Set Theory, Waltham, MA: Academic Press, 1978.

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