Theorem ax11f 1756

Description: Basis step for constructing a substitution instance of ax-11o 1588 without using ax-11o 1588. 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	. . 3 |- (ph -> A.xph)
2		ax-1 4	. . 3 |- (ph -> (x = y -> ph))
3	1, 2	19.21ai 1345	. 2 |- (ph -> A.x(x = y -> ph))
4	3	a1i12 9	1 |- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))

Syntax hints:  -. wn 2   -> wi 3  A.wal 1296   = wceq 1298

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-mp 7  ax-gen 1305  ax-4 1319  ax-5o 1321

Copyright terms: Public domain