Metamath Proof Explorer

Theorem eqss

Description: The subclass relationship is antisymmetric. Compare Theorem 4 of Suppes p. 22. (Contributed by NM, 21-May-1993)

Ref Expression
Assertion eqss ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 albiim ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ↔ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ∧ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴 ) ) )
2 dfcleq ⊢ ( 𝐴 = 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
3 dfss2 ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
4 dfss2 ⊢ ( 𝐵 ⊆ 𝐴 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴 ) )
5 3 4 anbi12i ⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) ↔ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ∧ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴 ) ) )
6 1 2 5 3bitr4i ⊢ ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) )