Theorem ax11f 1363

Description: Basis step for constructing a substitution instance of ax-11o 1216 without using ax-11o 1216. We can start with any formula ph in which x is not free.

Hypothesis

Ref	Expression
ax11f.1	|- (ph -> A.xph)

Assertion

Ref	Expression
ax11f	|- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))

Proof of Theorem ax11f

Step	Hyp	Ref	Expression
1		ax11f.1	. . . 4 |- (ph -> A.xph)
2		ax-1 4	. . . 4 |- (ph -> (x = y -> ph))
3	1, 2	19.21ai 996	. . 3 |- (ph -> A.x(x = y -> ph))
4	3	a1i 8	. 2 |- (x = y -> (ph -> A.x(x = y -> ph)))
5	4	a1i 8	1 |- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))

Syntax hints:  -. wn 2   -> wi 3  A.wal 952   = wceq 954

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-mp 7  ax-gen 961  ax-4 971  ax-5o 973

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