Theorem opeqpr 4231

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 2167	. 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 3757	. . 3 ⊢ ⟨𝐴, 𝐵⟩ = {{𝐴}, {𝐴, 𝐵}}
5	4	eqeq2i 2176	. 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 4164	. . 3 ⊢ {𝐴} ∈ V
9		prexg 4189	. . . 4 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ V) → {𝐴, 𝐵} ∈ V)
10	2, 3, 9	mp2an 423	. . 3 ⊢ {𝐴, 𝐵} ∈ V
11	6, 7, 8, 10	preq12b 3750	. 2 ⊢ ({𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))
12	1, 5, 11	3bitri 205	1 ⊢ (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))

Syntax hints:   ∧ wa 103   ↔ wb 104   ∨ wo 698   = wceq 1343   ∈ wcel 2136  Vcvv 2726  {csn 3576  {cpr 3577  ⟨cop 3579

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-14 2139  ax-ext 2147  ax-sep 4100  ax-pow 4153  ax-pr 4187

This theorem depends on definitions:  df-bi 116  df-3an 970  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-un 3120  df-in 3122  df-ss 3129  df-pw 3561  df-sn 3582  df-pr 3583  df-op 3585

This theorem is referenced by:  relop  4754