Theorem csbeq2

Description: Substituting into equivalent classes gives equivalent results. (Contributed by Giovanni Mascellani, 9-Apr-2018)

	Ref	Expression
Assertion	csbeq2	|- ( A. x B = C -> [_ A / x ]_ B = [_ A / x ]_ C )

Proof

Step	Hyp	Ref	Expression
1		eleq2	|- ( B = C -> ( y e. B <-> y e. C ) )
2	1	alimi	|- ( A. x B = C -> A. x ( y e. B <-> y e. C ) )
3		sbcbi2	|- ( A. x ( y e. B <-> y e. C ) -> ( [. A / x ]. y e. B <-> [. A / x ]. y e. C ) )
4	2 3	syl	|- ( A. x B = C -> ( [. A / x ]. y e. B <-> [. A / x ]. y e. C ) )
5	4	abbidv	|- ( A. x B = C -> { y | [. A / x ]. y e. B } = { y | [. A / x ]. y e. C } )
6		df-csb	|- [_ A / x ]_ B = { y | [. A / x ]. y e. B }
7		df-csb	|- [_ A / x ]_ C = { y | [. A / x ]. y e. C }
8	5 6 7	3eqtr4g	|- ( A. x B = C -> [_ A / x ]_ B = [_ A / x ]_ C )