Metamath Proof Explorer


Theorem equsex

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 . See equsexvw and equsexv for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsal . See equsexALT 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) (New usage is discouraged.)

Ref Expression
Hypotheses equsal.1
|- F/ x ps
equsal.2
|- ( x = y -> ( ph <-> ps ) )
Assertion equsex
|- ( E. x ( x = y /\ ph ) <-> ps )

Proof

Step Hyp Ref Expression
1 equsal.1
 |-  F/ x ps
2 equsal.2
 |-  ( x = y -> ( ph <-> ps ) )
3 2 biimpa
 |-  ( ( x = y /\ ph ) -> ps )
4 1 3 exlimi
 |-  ( E. x ( x = y /\ ph ) -> ps )
5 1 2 equsal
 |-  ( A. x ( x = y -> ph ) <-> ps )
6 equs4
 |-  ( A. x ( x = y -> ph ) -> E. x ( x = y /\ ph ) )
7 5 6 sylbir
 |-  ( ps -> E. x ( x = y /\ ph ) )
8 4 7 impbii
 |-  ( E. x ( x = y /\ ph ) <-> ps )