Axiom ax-c15

Description: Axiom ax-c15 was the original version of ax-12 , before it was discovered (in Jan. 2007) that the shorter ax-12 could replace it. It appears as Axiom scheme C15' in Megill p. 448 (p. 16 of the preprint). It is based on Lemma 16 of Tarski p. 70 and Axiom C8 of Monk2 p. 105, from which it can be proved by cases. To understand this theorem more easily, think of " -. A. x x = y -> ..." as informally meaning "if x and y are distinct variables then..." The antecedent becomes false if the same variable is substituted for x and y , ensuring the theorem is sound whenever this is the case. In some later theorems, we call an antecedent of the form -. A. x x = y a "distinctor."

Interestingly, if the wff expression substituted for ph contains no wff variables, the resulting statementcan be proved without invoking this axiom. This means that even though this axiom ismetalogically independent from the others, it is notlogically independent. Specifically, we can prove any wff-variable-free instance of axiom ax-c15 (from which the ax-12 instance follows by theorem ax12 .) The proof is by induction on formula length, using ax12eq and ax12el for the basis steps and ax12indn , ax12indi , and ax12inda for the induction steps. (This paragraph is true provided we use ax-c11 in place of ax-c11n .)

This axiom is obsolete and should no longer be used. It is proved above as theorem axc15 , which should be used instead. (Contributed by NM, 14-May-1993) (New usage is discouraged.)

	Ref	Expression
Assertion	ax-c15	$$⊢\neg \forall xx=y\to \left(x=y\to \left(\phi \to \forall x\left(x=y\to \phi \right)\right)\right)$$

Detailed syntax breakdown

Step	Hyp	Ref	Expression
0		vx	$$setvarx$$
1	0	cv	$$setvarx$$
2		vy	$$setvary$$
3	2	cv	$$setvary$$
4	1 3	wceq	$$wffx=y$$
5	4 0	wal	$$wff\forall xx=y$$
6	5	wn	$$wff\neg \forall xx=y$$
7		wph	$$wff\phi$$
8	4 7	wi	$$wff\left(x=y\to \phi \right)$$
9	8 0	wal	$$wff\forall x\left(x=y\to \phi \right)$$
10	7 9	wi	$$wff\left(\phi \to \forall x\left(x=y\to \phi \right)\right)$$
11	4 10	wi	$$wff\left(x=y\to \left(\phi \to \forall x\left(x=y\to \phi \right)\right)\right)$$
12	6 11	wi	$$wff\left(\neg \forall xx=y\to \left(x=y\to \left(\phi \to \forall x\left(x=y\to \phi \right)\right)\right)\right)$$