Theorem sbcbid

Description: Formula-building deduction for class substitution. (Contributed by NM, 29-Dec-2014)

	Ref	Expression
Hypotheses	sbcbid.1	|- F/ x ph
	sbcbid.2	|- ( ph -> ( ps <-> ch ) )
Assertion	sbcbid	|- ( ph -> ( [. A / x ]. ps <-> [. A / x ]. ch ) )

Proof

Step	Hyp	Ref	Expression
1		sbcbid.1	|- F/ x ph
2		sbcbid.2	|- ( ph -> ( ps <-> ch ) )
3	1 2	abbid	|- ( ph -> { x | ps } = { x | ch } )
4	3	eleq2d	|- ( ph -> ( A e. { x | ps } <-> A e. { x | ch } ) )
5		df-sbc	|- ( [. A / x ]. ps <-> A e. { x | ps } )
6		df-sbc	|- ( [. A / x ]. ch <-> A e. { x | ch } )
7	4 5 6	3bitr4g	|- ( ph -> ( [. A / x ]. ps <-> [. A / x ]. ch ) )