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Theorem opthpr 3787
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 3785 . 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 2361 . . . . . 6 (𝐴𝐷 ↔ ¬ 𝐴 = 𝐷)
8 pm2.21 618 . . . . . 6 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷))))
97, 8sylbi 121 . . . . 5 (𝐴𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷))))
109impd 254 . . . 4 (𝐴𝐷 → ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴 = 𝐶𝐵 = 𝐷)))
116, 10jaod 718 . . 3 (𝐴𝐷 → (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) → (𝐴 = 𝐶𝐵 = 𝐷)))
12 orc 713 . . 3 ((𝐴 = 𝐶𝐵 = 𝐷) → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
1311, 12impbid1 142 . 2 (𝐴𝐷 → (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
145, 13bitrid 192 1 (𝐴𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 104  wb 105  wo 709   = wceq 1364  wcel 2160  wne 2360  Vcvv 2752  {cpr 3608
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-in2 616  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2171
This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-nfc 2321  df-ne 2361  df-v 2754  df-un 3148  df-sn 3613  df-pr 3614
This theorem is referenced by: (None)
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