Metamath Proof Explorer


Theorem sbievw

Description: Conversion of implicit substitution to explicit substitution. Version of sbie and sbiev with more disjoint variable conditions, requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by BJ, 18-Jul-2023)

Ref Expression
Hypothesis sbievw.is
|- ( x = y -> ( ph <-> ps ) )
Assertion sbievw
|- ( [ y / x ] ph <-> ps )

Proof

Step Hyp Ref Expression
1 sbievw.is
 |-  ( x = y -> ( ph <-> ps ) )
2 sb6
 |-  ( [ y / x ] ph <-> A. x ( x = y -> ph ) )
3 1 equsalvw
 |-  ( A. x ( x = y -> ph ) <-> ps )
4 2 3 bitri
 |-  ( [ y / x ] ph <-> ps )