Axiom 3 Z(pos(Z) X(X intersection of Z P(X)))

Transcription

1 In this section, we are going to prove the equivalence between Axiom 3 ( the conjunction of any collection of positive properties is positive ) and Proposition 3 ( it is possible that God exists ). First, we have to translate these axiom/proposition which have been stated informally into formal ones. The following is the formal version of Axiom 3. Axiom 3 Z(pos(Z) X(X intersection of Z P(X))) Here, pos(z) is defined by X(Z(X) P(X)). This informally means that a property Z is applicable only to positive properties. On the other hand, X intersection of Z is defined by x(x(x) Y (Z(Y ) Y (x))). 1 This means, informally again, that a property X is necessarily applicable to objects which have properties contained in a property family Z. P(X) means, needless to say, that a property X is positive. The formal version of the Proposition 3 is as follows: Proposition 3 xg(x) Seeing that G(x) means that x is God, a further explanation except about the formal definition of G(x) would not be needed. A godness property G(x) is, using the lambda notation, defined by λx.( Y (P(Y ) Y (x))) which means an object which has all positive properties. Before proceeding to a proof that Axiom 3 and Proposition 3 above are actually equivalent, we have to formalize other informal axioms which are needed to prove the equivalence. First, we formalize Axiom 1. This axiom requires a property to be either positive or non-positive (that is, negative), but not both. There are two ways to formalize it. Axiom 1 A X(P( X) P(X)) B X( P(X) P( X)) In order to formalize Axiom 2 which asserts that any properties accompanied by a positive property are also positive, we need a new notation E(x) and the following abbreviations: E xφ is the abbreviation for x(e(x) Φ) and E xφ for x(e(x) Φ). 2 Using this new notation, we can formalize Axiom 2 as follows: Axiom 2 X Y ((P(X) E x(x(x) Y (x))) P(Y )) 3 1 This translation looks somewhat complicated because it deals with a property of properties. An analogy to Axiom of Union in axiomatic set theory X Y u(u Y z(z X u z)) might be helpful. 2 The rationale of this introduction of the new symbol E is concerning the issue about the two different interpretation of quantifiers in modal logic: actualist and possibilist. Although the issue is philosophically interesting, we set aside it in this paper. 3 The necessity symbol is needed here because Gödel thinks that the accompaniment of a property Y to Xis the nature of the property (Gödel, p. 403). 1

2 Now we are in a position to prove the equivalence. In proving, instead of directly showing the equivalence between Axiom 3 and Proposition 3, we prove it through the equivalences between the followings: 1. Axiom 3 2. P(G) 3. E xg(x) 4. Proposition 3 The derivability of P(G) from Axiom 3 would be intuitively clear because a property G is the collection of all positive properties. The formal proof is the following: Z(pos(Z) X(X intersection of Z P(X))) P(G) pos(p) X(X intersection of P P(X)) pos(p) X(X intersection of P P(X)) G intersection of P P(G) (G intersection of P) P(G) Given the definitions of pos and intersection of, both pos(p) and (G intersection of P) clearly lead to contradiction. However, we keep sticking to the formal proof here. Now pos(p) being an abbreviation of X(P(X) P(X)), pos(p) turns out to be: X(P(X) P(X)) X (P(X) P(X)) (P(P ) P(P )) P(P ) P(P ) In the same manner, G intersection of P and G being x(g(x) Y (P(Y ) Y (x))) and λx.( Y (P(Y ) Y (x))) respectively, the tree is going to be: 2

3 x(g(x) Y (P(Y ) Y (x))) x(g(x) Y (P(Y ) Y (x))) x (G(x) Y (P(Y ) Y (x))) w1 x (G(x) Y (P(Y ) Y (x))) w1 (G(a) Y (P(Y ) Y (a))) w1 (G(a) Y (P(Y ) Y (a))) w1 G(a) w1 Y (P(Y ) Y (a)) w1 Y (P(Y ) Y (a)) w1 (P(Q) Q(a)) w1 P(Q) w1 Q(a) w1 λx.( Y (P(Y ) Y (x)))(a) w1 Y (P(Y ) Y (a)) w1 P(Q) Q(a) w1 ( Y (P(Y ) Y (a)) G(a)) w1 Y (P(Y ) Y (a)) w1 G(a) w1 λx.( Y (P(Y ) Y (x)))(a) w1 Y (P(Y ) Y (a)) w1 Y (P(Y ) Y (a)) w1 (P(Q) Q(a)) w1 P(Q) w1 Q(a) w1 P(Q) Q(a) w1 P(Q) w1 w1 Q(a) w1 w1 P(Q) w1 w1 Q(a) w1 All branches considered, the above tree is closed. Therefore, it is proved that the Axiom 3 implies P(G). Next, we prove that P(G) implies E xg(x). In order to prove it, we first prove formally the following proposition the informal version of which we have already seen in the previous section. Proposition 1 X(P(X) E xx(x)) Roughly speaking, the informal version of Proposition 1 asserts that any positive properties can be instantiated. More exactly, it means that if a property X is positive, an object which has this property can exist. The formal expression of this assertion is X(P(X) E xx(x)). We need the following lemma first to prove formally Proposition 1. Lemma 1 XP(X) P(λx.x x) The validity of this lemma would be intuitively acceptable. If there exitst something positive, it must be identical to itself. The following tree is its proof (we assume Axiom 1 and Axiom 2). 3

4 X(P( X) P(X)) X Y ((P(X) E x(x(x) Y (x))) P(Y )) ( P(X) P(λx.x x)) XP(X) P(A) P(λx.x x) P(λx.x x) P( λx.x x) P(λx.x x) P( λx.x x) Y ((P(A) E x(a(x) Y (x))) P(Y )) (P(A) E x(a(x) λx.x = x(x))) P(λx.x = x) (continues to ) P(λx.x x) (P(A) E x(a(x) λx.x = x(x))) P(λx.x = x) P( λx.x = x) P(λx.x = x) E x(a(x) λx.x = x(x)) E x(a(x) λx.x = x(x)) x(e(x) (A(x) λx.x = x(x))) x (E(x) (A(x) λx.x = x(x))) w1 x (E(x) (A(x) λx.x = x(x))) w1 (E(a) (A(a) λx.x = x(a))) w1 E(a) w1 A(a) λx.x = x(a) w1 A(a) w1 λx.x = x(a) w1 λx.x x(a) w1 a a w1 P( λx.x = x) P(λx.x x) P(λx.x = x) With the above lemma and Axiom 2, we can prove Proposition 1 as follows. 4 4 In proving Proposition 1, I almost strictly follow Fitting because I could not find a more elegant proof than his proof (Fitting, p. 149). 4

5 X Y ((P(X) E x(x(x) Y (x))) P(Y )) XP(X) P(λx.x x) X(P(X) E xx(x)) X (P(X) E xx(x)) (P(A) E xa(x)) P(A) E xa(x) E xa(x) Y ((P(A) E x(a(x) Y (x))) P(Y )) (P(A) E x(a(x) λx.x x(x))) P(λx.x x) (P(A) E x(a(x) λx.x x(x))) P(λx.x x) E x(a(x) λx.x x(x)) E x(a(x) λx.x x(x)) w1 E x(a(x) λx.x x(x)) w1 x(e(x) (A(x) λx.x x(x))) w1 x (E(x) (A(x) λx.x x(x))) w1 (E(a) (A(a) λx.x x(a))) w1 E(a) w1 (A(a) λx.x x(a)) w1 A(a) w1 λx.x x(a) w1 E xa(x) w1 x(e(x) A(x)) w1 x (E(x) A(x)) w1 (E(a) A(a)) XP(X) X P(X) P(λx.x x) w1 E(a) w1 w1 A(a) w1 Once we get the above results, we can rather easily prove that P(G) implies E xg(x). P(G) X(P(X) E xx(x)) E xg(x) 5

6 P(G) E xg(x) P(G) E xg(x) Finally, we prove that E xg(x) implies Proposition 3. Seeing that E xg(x) is an abbreviation of (E(x) G(x)), the implication is intuitively straightforward. The following is the formal proof. E xg(x) xg(x) x(e(x) G(x)) xg(x) w1 x(e(x) G(x)) w1 E(a) G(a) w1 E(a) w1 G(a) w1 xg(x) w1 x G(x) w1 G(a) w1 All of the above taken into account, it has been formally shown that Axiom 3 (the recursiveness of any collection of positive properties) implies Proposition 3 (the possibility of the existence of God). To establish the equivalence between Axiom 3 and Proposition 3, next we move to the proof that Proposition 3 implies Axiom 3. First, we directly prove that Proposition 3 implies P(G) by introducing another definition of being God. The following is the definition. Another Definition of Godness G λx. Y (P(Y ) Y (x)) In addition to what the previous definition of Godness G says, this definition G also means that if something has some property, that property is positive. In fact, these two definition are equivalent. The following is its proof (we assume Axiom 1B). X( P(X) P( X)) x(g(x) G (x)) x (G(x) G (x)) (G(a) G (a)) (G(x) G (x)) (G (x) G(x)) First, we construct the left side of the tree. 6

7 (G(x) G (x)) G(x) G (x) λx. Y (P(Y ) Y (x))(a) Y (P(Y ) Y (a)) λx. Y (P(Y ) Y (x))(a) Y (P(Y ) Y (a)) Y (P(Y ) Y (a)) (P(A) A(a)) (P(A) A(a)) P(A) A(a) P(A) A(a) (A(a) P(A)) A(a) P( A) A(a) P(A) P( A) P( A) A(a) P( A) A(a) The next is the right side. (G (x) G(x)) λx. Y (P(Y ) Y (x))(a) Y (P(Y ) Y (a)) λx. Y (P(Y ) Y (x))(a) Y (P(Y ) Y (a)) Y (P(Y ) Y (a)) (P(A) A(a)) P(a) A(a) P(A) A(a) P(A) A(a) A(a) P(A) In addition to this equivalence x(g(x) G (x)), we need the formal version of Axiom 4 ( Any positive property is necessarily so, and any negative property is necessarily so ). 7

8 Axiom 4 A X(P(X) P(X)) B X( P(X) P(X)) 5 With the necessitated version of x(g(x) G (x)) 6 and Axiom 4, we can prove that G(x) implies P(G). xg(x) x(g(x) G (x)) X( P(X) P(X)) P(G) P(G) P(G) P(G) P(G) P(G) w1 xg(x) w1 G(a) w1 x(g(x) G (x)) w1 G(a) G (a) w1 G (a) w1 λx. Y (P(Y ) Y (x))(a) w1 Y (P(Y ) Y (a)) w1 P(G) G(a) w1 P(G) w1 P(G) w1 Now the only remaining task to prove the equivalence between Axiom 3 and Proposition 3 is to prove that P(G) implies Proposition 3. The proof is a bit lengthy. In the following proof, we assume Axiom 2. P(G) X Y ((P(X) E x(x(x) Y (x))) P(Y )) Z(pos(Z) X((X intersection of Z) P(X))) Z (pos(z) X((X intersection of Z) P(X))) (pos(a) X((X intersection of A) P(X))) pos(a) X(A(X) P(X)) X((X intersection of A) P(X)) X ((X intersection of A) P(X)) ((B intersection of A) P(B)) 5 In fact, Axiom 4B can be deducible from Axiom 4A by assuming Axiom 1. However, we omit its proof here. 6 An intuitive justification of x(g(x) G (x)) is that God has to be God in all possible worlds. Thought we can in fact deduce the necessitated version of x(g(x) G (x)), we omit the deduction. 8

9 B intersection of A x(b(x) Y (A(Y ) Y (x))) P(B) Y ((P(G) E x(g(x) Y (x))) P(Y )) (P(G) E x(g(x) B(x))) P(B) (P(G) E x(g(x) B(x))) P(B) P(G) E x(g(x) B(x)) E x(g(x) B(x)) x(e(x) (G(x) B(x))) x (E(x) (G(x) B(x))) w1 x (E(x) (G(x) B(x))) (E(a) (G(a) B(a))) E(a) (G(a) B(a)) G(a) B(a) x(b(x) Y (A(Y ) Y (x))) B(a) Y (A(Y ) Y (a)) B(a) Y (A(Y ) Y (a)) Y (A(Y ) Y (a)) B(a) (A(G) G(a)) B(a) (A(G) G(a)) A(G) G(a) B(a) 9