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Theorem opthpr 3584
Description: A way to represent ordered pairs using unordered pairs with distinct members. (Contributed by NM, 27-Mar-2007.)
Hypotheses
Ref Expression
preq12b.1 𝐴 ∈ V
preq12b.2 𝐵 ∈ V
preq12b.3 𝐶 ∈ V
preq12b.4 𝐷 ∈ V
Assertion
Ref Expression
opthpr (𝐴𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐵 = 𝐷)))

Proof of Theorem opthpr
StepHypRef Expression
1 preq12b.1 . . 3 𝐴 ∈ V
2 preq12b.2 . . 3 𝐵 ∈ V
3 preq12b.3 . . 3 𝐶 ∈ V
4 preq12b.4 . . 3 𝐷 ∈ V
51, 2, 3, 4preq12b 3582 . 2 ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
6 idd 21 . . . 4 (𝐴𝐷 → ((𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
7 df-ne 2250 . . . . . 6 (𝐴𝐷 ↔ ¬ 𝐴 = 𝐷)
8 pm2.21 580 . . . . . 6 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷))))
97, 8sylbi 119 . . . . 5 (𝐴𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷))))
109impd 251 . . . 4 (𝐴𝐷 → ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴 = 𝐶𝐵 = 𝐷)))
116, 10jaod 670 . . 3 (𝐴𝐷 → (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) → (𝐴 = 𝐶𝐵 = 𝐷)))
12 orc 666 . . 3 ((𝐴 = 𝐶𝐵 = 𝐷) → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
1311, 12impbid1 140 . 2 (𝐴𝐷 → (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
145, 13syl5bb 190 1 (𝐴𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 102  wb 103  wo 662   = wceq 1285  wcel 1434  wne 2249  Vcvv 2610  {cpr 3417
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065
This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-ne 2250  df-v 2612  df-un 2986  df-sn 3422  df-pr 3423
This theorem is referenced by: (None)
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