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Theorem opthpr 3694
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 3692 . 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 2307 . . . . . 6 (𝐴𝐷 ↔ ¬ 𝐴 = 𝐷)
8 pm2.21 606 . . . . . 6 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷))))
97, 8sylbi 120 . . . . 5 (𝐴𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷))))
109impd 252 . . . 4 (𝐴𝐷 → ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴 = 𝐶𝐵 = 𝐷)))
116, 10jaod 706 . . 3 (𝐴𝐷 → (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) → (𝐴 = 𝐶𝐵 = 𝐷)))
12 orc 701 . . 3 ((𝐴 = 𝐶𝐵 = 𝐷) → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
1311, 12impbid1 141 . 2 (𝐴𝐷 → (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
145, 13syl5bb 191 1 (𝐴𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 103  wb 104  wo 697   = wceq 1331  wcel 1480  wne 2306  Vcvv 2681  {cpr 3523
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-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119
This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-ne 2307  df-v 2683  df-un 3070  df-sn 3528  df-pr 3529
This theorem is referenced by: (None)
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