Theorem equs45f 1683

Description: Two ways of expressing substitution when y is not free in ph. (Contributed by NM, 25-Apr-2008.)

Hypothesis

Ref	Expression
equs45f.1	|- ( ph -> A. y ph )

Assertion

Ref	Expression
equs45f	|- ( E. x ( x = y /\ ph ) <-> A. x ( x = y -> ph ) )

Proof of Theorem equs45f

Step	Hyp	Ref	Expression
1		equs45f.1	. . . . 5 |- ( ph -> A. y ph )
2	1	anim2i 324	. . . 4 |- ( ( x = y /\ ph ) -> ( x = y /\ A. y ph ) )
3	2	eximi 1491	. . 3 |- ( E. x ( x = y /\ ph ) -> E. x ( x = y /\ A. y ph ) )
4		equs5a 1675	. . 3 |- ( E. x ( x = y /\ A. y ph ) -> A. x ( x = y -> ph ) )
5	3, 4	syl 14	. 2 |- ( E. x ( x = y /\ ph ) -> A. x ( x = y -> ph ) )
6		equs4 1613	. 2 |- ( A. x ( x = y -> ph ) -> E. x ( x = y /\ ph ) )
7	5, 6	impbii 117	1 |- ( E. x ( x = y /\ ph ) <-> A. x ( x = y -> ph ) )

Syntax hints:   -> wi 4   /\ wa 97   <-> wb 98   A.wal 1241   E.wex 1381

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 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-11 1397  ax-4 1400  ax-i9 1423  ax-ial 1427

This theorem depends on definitions:  df-bi 110

This theorem is referenced by:  sb5f  1685