Metamath Proof Explorer


Theorem bj-sbievw

Description: Lemma for substitution. Closed form of equsalvw and sbievw . (Contributed by BJ, 23-Jul-2023)

Ref Expression
Assertion bj-sbievw
|- ( [ y / x ] ( ph <-> ps ) -> ( [ y / x ] ph <-> ps ) )

Proof

Step Hyp Ref Expression
1 sb6
 |-  ( [ y / x ] ( ph <-> ps ) <-> A. x ( x = y -> ( ph <-> ps ) ) )
2 bj-sblem
 |-  ( A. x ( x = y -> ( ph <-> ps ) ) -> ( A. x ( x = y -> ph ) <-> ( E. x x = y -> ps ) ) )
3 sb6
 |-  ( [ y / x ] ph <-> A. x ( x = y -> ph ) )
4 ax6ev
 |-  E. x x = y
5 4 a1bi
 |-  ( ps <-> ( E. x x = y -> ps ) )
6 2 3 5 3bitr4g
 |-  ( A. x ( x = y -> ( ph <-> ps ) ) -> ( [ y / x ] ph <-> ps ) )
7 1 6 sylbi
 |-  ( [ y / x ] ( ph <-> ps ) -> ( [ y / x ] ph <-> ps ) )