Theorem opeqpr 5160

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 2806	. 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 4592	. . 3 ⊢ ⟨𝐴, 𝐵⟩ = {{𝐴}, {𝐴, 𝐵}}
5	4	eqeq2i 2811	. 2 ⊢ ({𝐶, 𝐷} = ⟨𝐴, 𝐵⟩ ↔ {𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}})
6		opeqpr.3	. . 3 ⊢ 𝐶 ∈ V
7		opeqpr.4	. . 3 ⊢ 𝐷 ∈ V
8		snex 5099	. . 3 ⊢ {𝐴} ∈ V
9		prex 5100	. . 3 ⊢ {𝐴, 𝐵} ∈ V
10	6, 7, 8, 9	preq12b 4567	. 2 ⊢ ({𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))
11	1, 5, 10	3bitri 289	1 ⊢ (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))

Syntax hints:   ↔ wb 198   ∧ wa 385   ∨ wo 874   = wceq 1653   ∈ wcel 2157  Vcvv 3385  {csn 4368  {cpr 4370  ⟨cop 4374

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777  ax-sep 4975  ax-nul 4983  ax-pr 5097

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-3an 1110  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-nfc 2930  df-v 3387  df-dif 3772  df-un 3774  df-in 3776  df-ss 3783  df-nul 4116  df-if 4278  df-sn 4369  df-pr 4371  df-op 4375

This theorem is referenced by:  propeqop  5163  relop  5476