Metamath Proof Explorer


Theorem sbrimvlem

Description: Common proof template for sbrimvw and sbrimv . The hypothesis is an instance of 19.21 . (Contributed by Wolf Lammen, 29-Jan-2024)

Ref Expression
Hypothesis sbrimvlem.1
|- ( A. x ( ph -> ( x = y -> ps ) ) <-> ( ph -> A. x ( x = y -> ps ) ) )
Assertion sbrimvlem
|- ( [ y / x ] ( ph -> ps ) <-> ( ph -> [ y / x ] ps ) )

Proof

Step Hyp Ref Expression
1 sbrimvlem.1
 |-  ( A. x ( ph -> ( x = y -> ps ) ) <-> ( ph -> A. x ( x = y -> ps ) ) )
2 sb6
 |-  ( [ y / x ] ( ph -> ps ) <-> A. x ( x = y -> ( ph -> ps ) ) )
3 bi2.04
 |-  ( ( ph -> ( x = y -> ps ) ) <-> ( x = y -> ( ph -> ps ) ) )
4 3 albii
 |-  ( A. x ( ph -> ( x = y -> ps ) ) <-> A. x ( x = y -> ( ph -> ps ) ) )
5 2 4 1 3bitr2i
 |-  ( [ y / x ] ( ph -> ps ) <-> ( ph -> A. x ( x = y -> ps ) ) )
6 sb6
 |-  ( [ y / x ] ps <-> A. x ( x = y -> ps ) )
7 6 imbi2i
 |-  ( ( ph -> [ y / x ] ps ) <-> ( ph -> A. x ( x = y -> ps ) ) )
8 5 7 bitr4i
 |-  ( [ y / x ] ( ph -> ps ) <-> ( ph -> [ y / x ] ps ) )