Theorem opprb 44525

Description: Equality for unordered pairs corresponds to equality of unordered pairs with the same elements. (Contributed by AV, 9-Jul-2023.)

Assertion

Ref	Expression
opprb	⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ∨ ⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩)))

Proof of Theorem opprb

Step	Hyp	Ref	Expression
1		preq12bg 4784	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))))
2		opthg 5392	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
3	2	adantr 481	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
4		opthg 5392	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
5	4	adantr 481	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
6	3, 5	orbi12d 916	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ((⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ∨ ⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩) ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))))
7	1, 6	bitr4d 281	1 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ∨ ⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396   ∨ wo 844   = wceq 1539   ∈ wcel 2106  {cpr 4563  ⟨cop 4567

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-dif 3890  df-un 3892  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568

This theorem is referenced by: (None)