Theorem opeqpr 4372

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

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 2236	. 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 3884	. . 3 ⊢ ⟨𝐴, 𝐵⟩ = {{𝐴}, {𝐴, 𝐵}}
5	4	eqeq2i 2245	. 2 ⊢ ({𝐶, 𝐷} = ⟨𝐴, 𝐵⟩ ↔ {𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}})
6		opeqpr.3	. . 3 ⊢ 𝐶 ∈ V
7		opeqpr.4	. . 3 ⊢ 𝐷 ∈ V
8	2	snex 4300	. . 3 ⊢ {𝐴} ∈ V
9		prexg 4327	. . . 4 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ V) → {𝐴, 𝐵} ∈ V)
10	2, 3, 9	mp2an 426	. . 3 ⊢ {𝐴, 𝐵} ∈ V
11	6, 7, 8, 10	preq12b 3876	. 2 ⊢ ({𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))
12	1, 5, 11	3bitri 206	1 ⊢ (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))

Syntax hints:   ∧ wa 104   ↔ wb 105   ∨ wo 716   = wceq 1398   ∈ wcel 2205  Vcvv 2815  {csn 3691  {cpr 3692  ⟨cop 3694

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-14 2208  ax-ext 2216  ax-sep 4230  ax-pow 4289  ax-pr 4324

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-v 2817  df-un 3217  df-in 3219  df-ss 3226  df-pw 3673  df-sn 3697  df-pr 3698  df-op 3700

This theorem is referenced by:  relop  4907