Theorem ax9o 1585

Description: An implication related to substitution. (Contributed by NM, 5-Aug-1993.) (Revised by NM, 3-Feb-2015.)

Assertion

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

Proof of Theorem ax9o

Step	Hyp	Ref	Expression
1		a9e 1583	. 2 |- E.x x = y
2		19.29r 1509	. . 3 |- ((E.x x = y /\ A.x(x = y -> A.xph)) -> E.x(x = y /\ (x = y -> A.xph)))
3		hba1 1430	. . . . 5 |- (A.xph -> A.xA.xph)
4		pm3.35 329	. . . . 5 |- ((x = y /\ (x = y -> A.xph)) -> A.xph)
5	3, 4	exlimih 1481	. . . 4 |- (E.x(x = y /\ (x = y -> A.xph)) -> A.xph)
6		ax-4 1397	. . . 4 |- (A.xph -> ph)
7	5, 6	syl 14	. . 3 |- (E.x(x = y /\ (x = y -> A.xph)) -> ph)
8	2, 7	syl 14	. 2 |- ((E.x x = y /\ A.x(x = y -> A.xph)) -> ph)
9	1, 8	mpan 400	1 |- (A.x(x = y -> A.xph) -> ph)

Syntax hints:   -> wi 4   /\ wa 97  A.wal 1240   = wceq 1242  E.wex 1378

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-4 1397  ax-i9 1420  ax-ial 1424

This theorem depends on definitions:  df-bi 110

This theorem is referenced by:  equsalh  1611  spimth  1620  spimh  1622