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Theorem brdifun 8527
Description: Evaluate the incomparability relation. (Contributed by Mario Carneiro, 9-Jul-2014.)
Hypothesis
Ref Expression
swoer.1 𝑅 = ((𝑋 × 𝑋) ∖ ( < < ))
Assertion
Ref Expression
brdifun ((𝐴𝑋𝐵𝑋) → (𝐴𝑅𝐵 ↔ ¬ (𝐴 < 𝐵𝐵 < 𝐴)))

Proof of Theorem brdifun
StepHypRef Expression
1 opelxpi 5626 . . . 4 ((𝐴𝑋𝐵𝑋) → ⟨𝐴, 𝐵⟩ ∈ (𝑋 × 𝑋))
2 df-br 5075 . . . 4 (𝐴(𝑋 × 𝑋)𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑋 × 𝑋))
31, 2sylibr 233 . . 3 ((𝐴𝑋𝐵𝑋) → 𝐴(𝑋 × 𝑋)𝐵)
4 swoer.1 . . . . . 6 𝑅 = ((𝑋 × 𝑋) ∖ ( < < ))
54breqi 5080 . . . . 5 (𝐴𝑅𝐵𝐴((𝑋 × 𝑋) ∖ ( < < ))𝐵)
6 brdif 5127 . . . . 5 (𝐴((𝑋 × 𝑋) ∖ ( < < ))𝐵 ↔ (𝐴(𝑋 × 𝑋)𝐵 ∧ ¬ 𝐴( < < )𝐵))
75, 6bitri 274 . . . 4 (𝐴𝑅𝐵 ↔ (𝐴(𝑋 × 𝑋)𝐵 ∧ ¬ 𝐴( < < )𝐵))
87baib 536 . . 3 (𝐴(𝑋 × 𝑋)𝐵 → (𝐴𝑅𝐵 ↔ ¬ 𝐴( < < )𝐵))
93, 8syl 17 . 2 ((𝐴𝑋𝐵𝑋) → (𝐴𝑅𝐵 ↔ ¬ 𝐴( < < )𝐵))
10 brun 5125 . . . 4 (𝐴( < < )𝐵 ↔ (𝐴 < 𝐵𝐴 < 𝐵))
11 brcnvg 5788 . . . . 5 ((𝐴𝑋𝐵𝑋) → (𝐴 < 𝐵𝐵 < 𝐴))
1211orbi2d 913 . . . 4 ((𝐴𝑋𝐵𝑋) → ((𝐴 < 𝐵𝐴 < 𝐵) ↔ (𝐴 < 𝐵𝐵 < 𝐴)))
1310, 12bitrid 282 . . 3 ((𝐴𝑋𝐵𝑋) → (𝐴( < < )𝐵 ↔ (𝐴 < 𝐵𝐵 < 𝐴)))
1413notbid 318 . 2 ((𝐴𝑋𝐵𝑋) → (¬ 𝐴( < < )𝐵 ↔ ¬ (𝐴 < 𝐵𝐵 < 𝐴)))
159, 14bitrd 278 1 ((𝐴𝑋𝐵𝑋) → (𝐴𝑅𝐵 ↔ ¬ (𝐴 < 𝐵𝐵 < 𝐴)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wa 396  wo 844   = wceq 1539  wcel 2106  cdif 3884  cun 3885  cop 4567   class class class wbr 5074   × cxp 5587  ccnv 5588
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3434  df-dif 3890  df-un 3892  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-br 5075  df-opab 5137  df-xp 5595  df-cnv 5597
This theorem is referenced by:  swoer  8528  swoord1  8529  swoord2  8530  swoso  8531
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