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 φAB
eqssd.2 φBA
Assertion eqssd φA=B

Proof

Step Hyp Ref Expression
1 eqssd.1 φAB
2 eqssd.2 φBA
3 eqss A=BABBA
4 1 2 3 sylanbrc φA=B