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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 ⊆ B
eqssi.2 ⊢ B ⊆ A
Assertion eqssi ⊢ A = B

Proof

Step Hyp Ref Expression
1 eqssi.1 ⊢ A ⊆ B
2 eqssi.2 ⊢ B ⊆ A
3 eqss ⊢ A = B ↔ A ⊆ B ∧ B ⊆ A
4 1 2 3 mpbir2an ⊢ A = B