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 )