Theorem eqelssd

Description: Equality deduction from subclass relationship and membership. (Contributed by AV, 21-Aug-2022)

	Ref	Expression
Hypotheses	eqelssd.1	⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
	eqelssd.2	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → 𝑥 ∈ 𝐴 )
Assertion	eqelssd	⊢ ( 𝜑 → 𝐴 = 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		eqelssd.1	⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
2		eqelssd.2	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → 𝑥 ∈ 𝐴 )
3	2	ex	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴 ) )
4	3	ssrdv	⊢ ( 𝜑 → 𝐵 ⊆ 𝐴 )
5	1 4	eqssd	⊢ ( 𝜑 → 𝐴 = 𝐵 )