Metamath Proof Explorer


Theorem eqri

Description: Infer equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 7-Oct-2017)

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

Proof

Step Hyp Ref Expression
1 eqri.1
 |-  F/_ x A
2 eqri.2
 |-  F/_ x B
3 eqri.3
 |-  ( x e. A <-> x e. B )
4 nftru
 |-  F/ x T.
5 3 a1i
 |-  ( T. -> ( x e. A <-> x e. B ) )
6 4 1 2 5 eqrd
 |-  ( T. -> A = B )
7 6 mptru
 |-  A = B