This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

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 )