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 ) )