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 
|- ( ph -> A C_ B )
eqssd.2 
|- ( ph -> B C_ A )
Assertion eqssd 
|- ( ph -> A = B )

Proof

Step Hyp Ref Expression
1 eqssd.1 
 |-  ( ph -> A C_ B )
2 eqssd.2 
 |-  ( ph -> B C_ A )
3 eqss 
 |-  ( A = B <-> ( A C_ B /\ B C_ A ) )
4 1 2 3 sylanbrc 
 |-  ( ph -> A = B )