Theorem cdeqab

Description: Distribute conditional equality over abstraction. (Contributed by Mario Carneiro, 11-Aug-2016)

	Ref	Expression
Hypothesis	cdeqnot.1	|- CondEq ( x = y -> ( ph <-> ps ) )
Assertion	cdeqab	|- CondEq ( x = y -> { z | ph } = { z | ps } )

Proof

Step	Hyp	Ref	Expression
1		cdeqnot.1	|- CondEq ( x = y -> ( ph <-> ps ) )
2	1	cdeqri	|- ( x = y -> ( ph <-> ps ) )
3	2	abbidv	|- ( x = y -> { z | ph } = { z | ps } )
4	3	cdeqi	|- CondEq ( x = y -> { z | ph } = { z | ps } )