Theorem eqrel2 39440

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 5768	. . 3 ⊢ (Rel 𝐴 → (𝐴 ⊆ 𝐵 ↔ ∀𝑥∀𝑦(𝑥𝐴𝑦 → 𝑥𝐵𝑦)))
2		ssrel3 5768	. . 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 3928	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		2albiim 1925	. 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 3900   class class class wbr 5106  Rel wrel 5661

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-ss 3901  df-br 5107  df-opab 5172  df-xp 5662  df-rel 5663

This theorem is used by: (None)