Theorem eqabrd

Description: Equality of a class variable and a class abstraction (deduction form of eqabb ). (Contributed by NM, 16-Nov-1995)

	Ref	Expression
Hypothesis	eqabrd.1	|- ( ph -> A = { x | ps } )
Assertion	eqabrd	|- ( ph -> ( x e. A <-> ps ) )

Proof

Step	Hyp	Ref	Expression
1		eqabrd.1	|- ( ph -> A = { x | ps } )
2	1	eleq2d	|- ( ph -> ( x e. A <-> x e. { x | ps } ) )
3		abid	|- ( x e. { x | ps } <-> ps )
4	2 3	bitrdi	|- ( ph -> ( x e. A <-> ps ) )