Theorem opeqpr 5449

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 2748	. 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 4806	. . 3 ⊢ ⟨𝐴, 𝐵⟩ = {{𝐴}, {𝐴, 𝐵}}
5	4	eqeq2i 2754	. 2 ⊢ ({𝐶, 𝐷} = ⟨𝐴, 𝐵⟩ ↔ {𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}})
6		opeqpr.3	. . 3 ⊢ 𝐶 ∈ V
7		opeqpr.4	. . 3 ⊢ 𝐷 ∈ V
8		snex 5371	. . 3 ⊢ {𝐴} ∈ V
9		prex 5370	. . 3 ⊢ {𝐴, 𝐵} ∈ V
10	6, 7, 8, 9	preq12b 4784	. 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 397   ∨ wo 854   = wceq 1548   ∈ wcel 2121  Vcvv 3433  {csn 4558  {cpr 4560  ⟨cop 4564

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-8 2123  ax-9 2131  ax-ext 2713  ax-sep 5221  ax-pr 5365

This theorem depends on definitions:  df-bi 209  df-an 398  df-or 855  df-3an 1095  df-tru 1551  df-fal 1561  df-ex 1788  df-sb 2075  df-clab 2720  df-cleq 2733  df-clel 2816  df-v 3435  df-dif 3888  df-un 3890  df-ss 3902  df-nul 4265  df-if 4458  df-sn 4559  df-pr 4561  df-op 4565

This theorem is referenced by:  propeqop  5451  relop  5795