Metamath Proof Explorer


Theorem equsalhw

Description: Version of equsalh with a disjoint variable condition, which does not require ax-13 . (Contributed by NM, 29-Nov-2015) (Proof shortened by Wolf Lammen, 8-Jul-2022)

Ref Expression
Hypotheses equsalhw.1
|- ( ps -> A. x ps )
equsalhw.2
|- ( x = y -> ( ph <-> ps ) )
Assertion equsalhw
|- ( A. x ( x = y -> ph ) <-> ps )

Proof

Step Hyp Ref Expression
1 equsalhw.1
 |-  ( ps -> A. x ps )
2 equsalhw.2
 |-  ( x = y -> ( ph <-> ps ) )
3 1 nf5i
 |-  F/ x ps
4 3 2 equsalv
 |-  ( A. x ( x = y -> ph ) <-> ps )