Metamath Proof Explorer


Theorem eqelssd

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

Ref Expression
Hypotheses eqelssd.1 φAB
eqelssd.2 φxBxA
Assertion eqelssd φA=B

Proof

Step Hyp Ref Expression
1 eqelssd.1 φAB
2 eqelssd.2 φxBxA
3 2 ex φxBxA
4 3 ssrdv φBA
5 1 4 eqssd φA=B