Metamath Proof Explorer


Theorem eqelssd

Description: Equality deduction from subclass relationship and membership. (Contributed by AV, 21-Aug-2022)

Ref Expression
Hypotheses eqelssd.1 φ A B
eqelssd.2 φ x B x A
Assertion eqelssd φ A = B

Proof

Step Hyp Ref Expression
1 eqelssd.1 φ A B
2 eqelssd.2 φ x B x A
3 2 ex φ x B x A
4 3 ssrdv φ B A
5 1 4 eqssd φ A = B