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Metamath Proof Explorer

Theorem equsal

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 . See equsalvw and equsalv for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsex . (Contributed by NM, 2-Jun-1993) (Proof shortened by Andrew Salmon, 12-Aug-2011) (Revised by Mario Carneiro, 3-Oct-2016) (Proof shortened by Wolf Lammen, 5-Feb-2018) (New usage is discouraged.)

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

Proof

Step Hyp Ref Expression
1 equsal.1 
 |-  F/ x ps
2 equsal.2 
 |-  ( x = y -> ( ph <-> ps ) )
3 1 19.23 
 |-  ( A. x ( x = y -> ps ) <-> ( E. x x = y -> ps ) )
4 2 pm5.74i 
 |-  ( ( x = y -> ph ) <-> ( x = y -> ps ) )
5 4 albii 
 |-  ( A. x ( x = y -> ph ) <-> A. x ( x = y -> ps ) )
6 ax6e 
 |-  E. x x = y
7 6 a1bi 
 |-  ( ps <-> ( E. x x = y -> ps ) )
8 3 5 7 3bitr4i 
 |-  ( A. x ( x = y -> ph ) <-> ps )