Theorem eqrd

Description: Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017) (Proof shortened by BJ, 1-Dec-2021)

	Ref	Expression
Hypotheses	eqrd.0	|- F/ x ph
	eqrd.1	|- F/_ x A
	eqrd.2	|- F/_ x B
	eqrd.3	|- ( ph -> ( x e. A <-> x e. B ) )
Assertion	eqrd	|- ( ph -> A = B )

Proof

Step	Hyp	Ref	Expression
1		eqrd.0	|- F/ x ph
2		eqrd.1	|- F/_ x A
3		eqrd.2	|- F/_ x B
4		eqrd.3	|- ( ph -> ( x e. A <-> x e. B ) )
5	1 4	alrimi	|- ( ph -> A. x ( x e. A <-> x e. B ) )
6	2 3	cleqf	|- ( A = B <-> A. x ( x e. A <-> x e. B ) )
7	5 6	sylibr	|- ( ph -> A = B )