Theorem equsex 2129

Description: An equivalence related to implicit substitution. See equsexv 2065 for a version with a dv condition which does not require ax-13 2090. See equsexALT 2130 for an alternate proof. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 3-Oct-2016.) (Proof shortened by Wolf Lammen, 6-Feb-2018.)

Hypotheses

Ref	Expression
equsex.nf	|- F/ x ps
equsex.1	|- ( x = y -> ( ph <-> ps ) )

Assertion

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

Proof of Theorem equsex

Step	Hyp	Ref	Expression
1		equsex.nf	. . 3 |- F/ x ps
2		equsex.1	. . . 4 |- ( x = y -> ( ph <-> ps ) )
3	2	biimpa 487	. . 3 |- ( ( x = y /\ ph ) -> ps )
4	1, 3	exlimi 1994	. 2 |- ( E. x ( x = y /\ ph ) -> ps )
5	1, 2	equsal 2127	. . 3 |- ( A. x ( x = y -> ph ) <-> ps )
6		equs4 2126	. . 3 |- ( A. x ( x = y -> ph ) -> E. x ( x = y /\ ph ) )
7	5, 6	sylbir 217	. 2 |- ( ps -> E. x ( x = y /\ ph ) )
8	4, 7	impbii 191	1 |- ( E. x ( x = y /\ ph ) <-> ps )

Syntax hints:   -> wi 4   <-> wb 188   /\ wa 371   A.wal 1441   E.wex 1662   F/wnf 1666

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1668  ax-4 1681  ax-5 1757  ax-6 1804  ax-7 1850  ax-10 1914  ax-12 1932  ax-13 2090

This theorem depends on definitions:  df-bi 189  df-an 373  df-ex 1663  df-nf 1667

This theorem is referenced by:  equsexh  2131  sb5rf  2250