Metamath Proof Explorer


Theorem eqssd

Description: Equality deduction from two subclass relationships. Compare Theorem 4 of Suppes p. 22. (Contributed by NM, 27-Jun-2004)

Ref Expression
Hypotheses eqssd.1 φ A B
eqssd.2 φ B A
Assertion eqssd φ A = B

Proof

Step Hyp Ref Expression
1 eqssd.1 φ A B
2 eqssd.2 φ B A
3 eqss A = B A B B A
4 1 2 3 sylanbrc φ A = B