Metamath Proof Explorer


Theorem eqrdv

Description: Deduce equality of classes from equivalence of membership. (Contributed by NM, 17-Mar-1996)

Ref Expression
Hypothesis eqrdv.1
|- ( ph -> ( x e. A <-> x e. B ) )
Assertion eqrdv
|- ( ph -> A = B )

Proof

Step Hyp Ref Expression
1 eqrdv.1
 |-  ( ph -> ( x e. A <-> x e. B ) )
2 1 alrimiv
 |-  ( ph -> A. x ( x e. A <-> x e. B ) )
3 dfcleq
 |-  ( A = B <-> A. x ( x e. A <-> x e. B ) )
4 2 3 sylibr
 |-  ( ph -> A = B )