Theorem eqrel2 39331

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 5761	. . 3 ⊢ (Rel 𝐴 → (𝐴 ⊆ 𝐵 ↔ ∀𝑥∀𝑦(𝑥𝐴𝑦 → 𝑥𝐵𝑦)))
2		ssrel3 5761	. . 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 3923	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		2albiim 1922	. 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 1567   = wceq 1569   ⊆ wss 3895   class class class wbr 5102  Rel wrel 5655

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-ss 3896  df-br 5103  df-opab 5167  df-xp 5656  df-rel 5657

This theorem is used by: (None)