Theorem opprb 47906

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 4816	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))))
2		opthg 5457	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
3	2	adantr 486	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
4		opthg 5457	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
5	4	adantr 486	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
6	3, 5	orbi12d 932	. 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 285	1 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ∨ ⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570   ∈ wcel 2145  {cpr 4589  ⟨cop 4593

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-sep 5255  ax-pr 5402

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594

This theorem is used by: (None)