Theorem cdeqab1

Description: Distribute conditional equality over abstraction. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by Mario Carneiro, 11-Aug-2016) (New usage is discouraged.)

	Ref	Expression
Hypothesis	cdeqnot.1	|- CondEq ( x = y -> ( ph <-> ps ) )
Assertion	cdeqab1	|- CondEq ( x = y -> { x | ph } = { y | ps } )

Proof

Step	Hyp	Ref	Expression
1		cdeqnot.1	|- CondEq ( x = y -> ( ph <-> ps ) )
2		nfv	|- F/ y ph
3		nfv	|- F/ x ps
4	1	cdeqri	|- ( x = y -> ( ph <-> ps ) )
5	2 3 4	cbvab	|- { x | ph } = { y | ps }
6	5	cdeqth	|- CondEq ( x = y -> { x | ph } = { y | ps } )