Theorem eqrel2 39205

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

Assertion

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

Distinct variable groups:   𝑥,𝐴,𝑦   𝑥,𝐵,𝑦

Proof of Theorem eqrel2

Step	Hyp	Ref	Expression
1		ssrel3 5762	. . 3 ⊢ (Rel 𝐴 → (𝐴 ⊆ 𝐵 ↔ ∀𝑥∀𝑦(𝑥𝐴𝑦 → 𝑥𝐵𝑦)))
2		ssrel3 5762	. . 3 ⊢ (Rel 𝐵 → (𝐵 ⊆ 𝐴 ↔ ∀𝑥∀𝑦(𝑥𝐵𝑦 → 𝑥𝐴𝑦)))
3	1, 2	bi2anan9 650	. 2 ⊢ ((Rel 𝐴 ∧ Rel 𝐵) → ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) ↔ (∀𝑥∀𝑦(𝑥𝐴𝑦 → 𝑥𝐵𝑦) ∧ ∀𝑥∀𝑦(𝑥𝐵𝑦 → 𝑥𝐴𝑦))))
4		eqss 3946	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		2albiim 1923	. 2 ⊢ (∀𝑥∀𝑦(𝑥𝐴𝑦 ↔ 𝑥𝐵𝑦) ↔ (∀𝑥∀𝑦(𝑥𝐴𝑦 → 𝑥𝐵𝑦) ∧ ∀𝑥∀𝑦(𝑥𝐵𝑦 → 𝑥𝐴𝑦)))
6	3, 4, 5	3bitr4g 317	1 ⊢ ((Rel 𝐴 ∧ Rel 𝐵) → (𝐴 = 𝐵 ↔ ∀𝑥∀𝑦(𝑥𝐴𝑦 ↔ 𝑥𝐵𝑦)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570   ⊆ wss 3899   class class class wbr 5103  Rel wrel 5656

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3453  df-ss 3916  df-br 5104  df-opab 5168  df-xp 5657  df-rel 5658

This theorem is used by: (None)