Metamath Proof Explorer


Theorem sbiev

Description: Conversion of implicit substitution to explicit substitution. Version of sbie with a disjoint variable condition, not requiring ax-13 . See sbievw for a version with a disjoint variable condition requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by Wolf Lammen, 18-Jan-2023) Remove dependence on ax-10 and shorten proof. (Revised by BJ, 18-Jul-2023) (Proof shortened by SN, 24-Jul-2025)

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

Proof

Step Hyp Ref Expression
1 sbiev.1
 |-  F/ x ps
2 sbiev.2
 |-  ( x = y -> ( ph <-> ps ) )
3 2 sbbiiev
 |-  ( [ y / x ] ph <-> [ y / x ] ps )
4 1 sbf
 |-  ( [ y / x ] ps <-> ps )
5 3 4 bitri
 |-  ( [ y / x ] ph <-> ps )