Theorem opprb 43339

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 4770	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))))
2		opthg 5355	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
3	2	adantr 483	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
4		opthg 5355	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
5	4	adantr 483	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
6	3, 5	orbi12d 915	. 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 284	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 208   ∧ wa 398   ∨ wo 843   = wceq 1537   ∈ wcel 2114  {cpr 4555  ⟨cop 4559

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2793  ax-sep 5189  ax-nul 5196  ax-pr 5316

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-rab 3147  df-v 3488  df-dif 3927  df-un 3929  df-in 3931  df-ss 3940  df-nul 4280  df-if 4454  df-sn 4554  df-pr 4556  df-op 4560

This theorem is referenced by: (None)