Theorem ax9 1110

Description: A variant of ax-9 1102. Axiom scheme C10' in [Megill] p. 448 (p. 16 of the preprint). Theorem ax9a 1111 shows that ax-9 1102 and ax9 1110 are equivalent in the presence of the others.

Assertion

Ref	Expression
ax9	|- (A.x(x = y -> A.xph) -> ph)

Proof of Theorem ax9

Step	Hyp	Ref	Expression
1		ax-9 1102	. . . 4 |- -. A.x -. x = y
2		df-ex 957	. . . 4 |- (E.x x = y <-> -. A.x -. x = y)
3	1, 2	mpbir 190	. . 3 |- E.x x = y
4		19.22 1015	. . 3 |- (A.x(x = y -> A.xph) -> (E.x x = y -> E.xA.xph))
5	3, 4	mpi 44	. 2 |- (A.x(x = y -> A.xph) -> E.xA.xph)
6		a6e 966	. 2 |- (E.xA.xph -> ph)
7	5, 6	syl 10	1 |- (A.x(x = y -> A.xph) -> ph)

Syntax hints:  -. wn 2   -> wi 3  A.wal 950  E.wex 956   = wceq 1099

This theorem is referenced by:  ax9a 1111  equid 1113  equs4 1133  equsal 1134  a4at 1141  a4a 1142  cbv1 1145

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-gen 955  ax-9 1102

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 957

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