Theorem opeqpr 5476

Description: Equivalence for an ordered pair equal to an unordered pair. (Contributed by NM, 3-Jun-2008.) (Avoid depending on this detail.)

Hypotheses

Ref	Expression
opeqpr.1	⊢ 𝐴 ∈ V
opeqpr.2	⊢ 𝐵 ∈ V
opeqpr.3	⊢ 𝐶 ∈ V
opeqpr.4	⊢ 𝐷 ∈ V

Assertion

Ref	Expression
opeqpr	⊢ (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))

Proof of Theorem opeqpr

Step	Hyp	Ref	Expression
1		eqcom 2771	. 2 ⊢ (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ {𝐶, 𝐷} = ⟨𝐴, 𝐵⟩)
2		opeqpr.1	. . . 4 ⊢ 𝐴 ∈ V
3		opeqpr.2	. . . 4 ⊢ 𝐵 ∈ V
4	2, 3	dfop 4832	. . 3 ⊢ ⟨𝐴, 𝐵⟩ = {{𝐴}, {𝐴, 𝐵}}
5	4	eqeq2i 2777	. 2 ⊢ ({𝐶, 𝐷} = ⟨𝐴, 𝐵⟩ ↔ {𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}})
6		opeqpr.3	. . 3 ⊢ 𝐶 ∈ V
7		opeqpr.4	. . 3 ⊢ 𝐷 ∈ V
8		snex 5398	. . 3 ⊢ {𝐴} ∈ V
9		prex 5397	. . 3 ⊢ {𝐴, 𝐵} ∈ V
10	6, 7, 8, 9	preq12b 4810	. 2 ⊢ ({𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))
11	1, 5, 10	3bitri 299	1 ⊢ (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))

Syntax hints:   ↔ wb 208   ∧ wa 399   ∨ wo 858   = wceq 1562   ∈ wcel 2144  Vcvv 3456  {csn 4584  {cpr 4586  ⟨cop 4590

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-ext 2736  ax-sep 5248  ax-pr 5392

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1101  df-tru 1565  df-fal 1575  df-ex 1802  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-v 3458  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4288  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591

This theorem is referenced by:  propeqop  5478  relop  5824