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Metamath Proof Explorer

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 )