Theorem opeqpr 5504

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 2740	. 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 4871	. . 3 ⊢ ⟨𝐴, 𝐵⟩ = {{𝐴}, {𝐴, 𝐵}}
5	4	eqeq2i 2746	. 2 ⊢ ({𝐶, 𝐷} = ⟨𝐴, 𝐵⟩ ↔ {𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}})
6		opeqpr.3	. . 3 ⊢ 𝐶 ∈ V
7		opeqpr.4	. . 3 ⊢ 𝐷 ∈ V
8		snex 5430	. . 3 ⊢ {𝐴} ∈ V
9		prex 5431	. . 3 ⊢ {𝐴, 𝐵} ∈ V
10	6, 7, 8, 9	preq12b 4850	. 2 ⊢ ({𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))
11	1, 5, 10	3bitri 297	1 ⊢ (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))

Syntax hints:   ↔ wb 205   ∧ wa 397   ∨ wo 846   = wceq 1542   ∈ wcel 2107  Vcvv 3475  {csn 4627  {cpr 4629  ⟨cop 4633

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704  ax-sep 5298  ax-nul 5305  ax-pr 5426

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-3an 1090  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3477  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964  df-nul 4322  df-if 4528  df-sn 4628  df-pr 4630  df-op 4634

This theorem is referenced by:  propeqop  5506  relop  5848