Theorem 2sbbid

Description: Deduction doubly substituting both sides of a biconditional. (Contributed by AV, 30-Jul-2023)

	Ref	Expression
Hypotheses	sbbid.1	|- F/ x ph
	sbbid.2	|- ( ph -> ( ps <-> ch ) )
	2sbbid.1	|- F/ y ph
Assertion	2sbbid	|- ( ph -> ( [ t / x ] [ u / y ] ps <-> [ t / x ] [ u / y ] ch ) )

Proof

Step	Hyp	Ref	Expression
1		sbbid.1	|- F/ x ph
2		sbbid.2	|- ( ph -> ( ps <-> ch ) )
3		2sbbid.1	|- F/ y ph
4	3 2	sbbid	|- ( ph -> ( [ u / y ] ps <-> [ u / y ] ch ) )
5	1 4	sbbid	|- ( ph -> ( [ t / x ] [ u / y ] ps <-> [ t / x ] [ u / y ] ch ) )