Theorem opprb 46413

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 4855	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))))
2		opthg 5479	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
3	2	adantr 480	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
4		opthg 5479	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
5	4	adantr 480	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
6	3, 5	orbi12d 917	. 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 282	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 395   ∨ wo 846   = wceq 1534   ∈ wcel 2099  {cpr 4631  ⟨cop 4635

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699  ax-sep 5299  ax-nul 5306  ax-pr 5429

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-rab 3430  df-v 3473  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964  df-nul 4324  df-if 4530  df-sn 4630  df-pr 4632  df-op 4636

This theorem is referenced by: (None)