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		df-ss	⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
4		df-ss	⊢ ( 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 ) )