Metamath Proof Explorer


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 xφ
eqrd.1 _xA
eqrd.2 _xB
eqrd.3 φxAxB
Assertion eqrd φA=B

Proof

Step Hyp Ref Expression
1 eqrd.0 xφ
2 eqrd.1 _xA
3 eqrd.2 _xB
4 eqrd.3 φxAxB
5 1 4 alrimi φxxAxB
6 2 3 cleqf A=BxxAxB
7 5 6 sylibr φA=B