Theorem ax9o 1480

Description: Show that the original axiom ax-9o 1481 can be derived from ax-9 1307 and others. See ax9 1482 for the rederivation of ax-9 1307 from ax-9o 1481.

This theorem should not be referenced in any proof. Instead, use ax-9o 1481 below so that uses of ax-9o 1481 can be more easily identified.

Assertion

Ref	Expression
ax9o	|- (A.x(x = y -> A.xph) -> ph)

Proof of Theorem ax9o

Step	Hyp	Ref	Expression
1		ax-9 1307	. . . 4 |- -. A.x -. x = y
2		df-ex 1327	. . . 4 |- (E.x x = y <-> -. A.x -. x = y)
3	1, 2	mpbir 207	. . 3 |- E.x x = y
4		exim 1386	. . 3 |- (A.x(x = y -> A.xph) -> (E.x x = y -> E.xA.xph))
5	3, 4	mpi 55	. 2 |- (A.x(x = y -> A.xph) -> E.xA.xph)
6		a6e 1336	. 2 |- (E.xA.xph -> ph)
7	5, 6	syl 12	1 |- (A.x(x = y -> A.xph) -> ph)

Syntax hints:  -. wn 2   -> wi 3  A.wal 1296   = wceq 1298  E.wex 1326

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 1305  ax-9 1307  ax-4 1319  ax-5o 1321  ax-6o 1324

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327

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