Metamath Proof Explorer

Theorem eqssi

Description: Infer equality from two subclass relationships. Compare Theorem 4 of Suppes p. 22. (Contributed by NM, 9-Sep-1993)

Ref Expression
Hypotheses eqssi.1 
|- A C_ B
eqssi.2 
|- B C_ A
Assertion eqssi 
|- A = B

Proof

Step Hyp Ref Expression
1 eqssi.1 
 |-  A C_ B
2 eqssi.2 
 |-  B C_ A
3 eqss 
 |-  ( A = B <-> ( A C_ B /\ B C_ A ) )
4 1 2 3 mpbir2an 
 |-  A = B