Theorem eqrel2

Description: Equality of relations. (Contributed by Peter Mazsa, 8-Mar-2019)

	Ref	Expression
Assertion	eqrel2	⊢ ( ( Rel 𝐴 ∧ Rel 𝐵 ) → ( 𝐴 = 𝐵 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 ↔ 𝑥 𝐵 𝑦 ) ) )

Proof

Step	Hyp	Ref	Expression
1		ssrel3	⊢ ( Rel 𝐴 → ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 → 𝑥 𝐵 𝑦 ) ) )
2		ssrel3	⊢ ( Rel 𝐵 → ( 𝐵 ⊆ 𝐴 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐵 𝑦 → 𝑥 𝐴 𝑦 ) ) )
3	1 2	bi2anan9	⊢ ( ( Rel 𝐴 ∧ Rel 𝐵 ) → ( ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) ↔ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 → 𝑥 𝐵 𝑦 ) ∧ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐵 𝑦 → 𝑥 𝐴 𝑦 ) ) ) )
4		eqss	⊢ ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) )
5		2albiim	⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 ↔ 𝑥 𝐵 𝑦 ) ↔ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 → 𝑥 𝐵 𝑦 ) ∧ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐵 𝑦 → 𝑥 𝐴 𝑦 ) ) )
6	3 4 5	3bitr4g	⊢ ( ( Rel 𝐴 ∧ Rel 𝐵 ) → ( 𝐴 = 𝐵 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝐴 𝑦 ↔ 𝑥 𝐵 𝑦 ) ) )