Theorem opprb 47713

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 4817	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))))
2		opthg 5459	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
3	2	adantr 485	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
4		opthg 5459	. . . 4 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
5	4	adantr 485	. . 3 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ (𝐶 ∈ 𝑋 ∧ 𝐷 ∈ 𝑌)) → (⟨𝐴, 𝐵⟩ = ⟨𝐷, 𝐶⟩ ↔ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
6	3, 5	orbi12d 931	. 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⟩)))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   ∨ wo 860   = wceq 1568   ∈ wcel 2141  {cpr 4590  ⟨cop 4594

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2143  ax-9 2151  ax-ext 2733  ax-sep 5256  ax-pr 5404

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1571  df-fal 1581  df-ex 1808  df-sb 2095  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3415  df-v 3455  df-dif 3907  df-un 3909  df-ss 3921  df-nul 4286  df-if 4487  df-sn 4589  df-pr 4591  df-op 4595

This theorem is referenced by: (None)