Metamath Proof Explorer


Theorem sbbidv

Description: Deduction substituting both sides of a biconditional, with ph and x disjoint. See also sbbid . (Contributed by Wolf Lammen, 6-May-2023) (Proof shortened by Steven Nguyen, 6-Jul-2023)

Ref Expression
Hypothesis sbbidv.1
|- ( ph -> ( ps <-> ch ) )
Assertion sbbidv
|- ( ph -> ( [ t / x ] ps <-> [ t / x ] ch ) )

Proof

Step Hyp Ref Expression
1 sbbidv.1
 |-  ( ph -> ( ps <-> ch ) )
2 1 alrimiv
 |-  ( ph -> A. x ( ps <-> ch ) )
3 spsbbi
 |-  ( A. x ( ps <-> ch ) -> ( [ t / x ] ps <-> [ t / x ] ch ) )
4 2 3 syl
 |-  ( ph -> ( [ t / x ] ps <-> [ t / x ] ch ) )