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Theorem opthpr 3570
 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 3568 . 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 2221 . . . . . 6 (𝐴𝐷 ↔ ¬ 𝐴 = 𝐷)
8 pm2.21 557 . . . . . 6 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷))))
97, 8sylbi 118 . . . . 5 (𝐴𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷))))
109impd 246 . . . 4 (𝐴𝐷 → ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴 = 𝐶𝐵 = 𝐷)))
116, 10jaod 647 . . 3 (𝐴𝐷 → (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) → (𝐴 = 𝐶𝐵 = 𝐷)))
12 orc 643 . . 3 ((𝐴 = 𝐶𝐵 = 𝐷) → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
1311, 12impbid1 134 . 2 (𝐴𝐷 → (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
145, 13syl5bb 185 1 (𝐴𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 101   ↔ wb 102   ∨ wo 639   = wceq 1259   ∈ wcel 1409   ≠ wne 2220  Vcvv 2574  {cpr 3403 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in2 555  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038 This theorem depends on definitions:  df-bi 114  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-ne 2221  df-v 2576  df-un 2949  df-sn 3408  df-pr 3409 This theorem is referenced by: (None)
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