Metamath Proof Explorer


Theorem sbequi

Description: An equality theorem for substitution. (Contributed by NM, 14-May-1993) (Proof shortened by Wolf Lammen, 15-Sep-2018) (Proof shortened by Steven Nguyen, 7-Jul-2023)

Ref Expression
Assertion sbequi
|- ( x = y -> ( [ x / z ] ph -> [ y / z ] ph ) )

Proof

Step Hyp Ref Expression
1 sbequ
 |-  ( x = y -> ( [ x / z ] ph <-> [ y / z ] ph ) )
2 1 biimpd
 |-  ( x = y -> ( [ x / z ] ph -> [ y / z ] ph ) )