Metamath Proof Explorer


Theorem sbimdv

Description: Deduction substituting both sides of an implication, with ph and x disjoint. See also sbimd . (Contributed by Wolf Lammen, 6-May-2023) Revise df-sb . (Revised by Steven Nguyen, 6-Jul-2023)

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

Proof

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