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Theorem brdifun 8775
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 5722 . . . 4 ((𝐴𝑋𝐵𝑋) → ⟨𝐴, 𝐵⟩ ∈ (𝑋 × 𝑋))
2 df-br 5144 . . . 4 (𝐴(𝑋 × 𝑋)𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑋 × 𝑋))
31, 2sylibr 234 . . 3 ((𝐴𝑋𝐵𝑋) → 𝐴(𝑋 × 𝑋)𝐵)
4 swoer.1 . . . . . 6 𝑅 = ((𝑋 × 𝑋) ∖ ( < < ))
54breqi 5149 . . . . 5 (𝐴𝑅𝐵𝐴((𝑋 × 𝑋) ∖ ( < < ))𝐵)
6 brdif 5196 . . . . 5 (𝐴((𝑋 × 𝑋) ∖ ( < < ))𝐵 ↔ (𝐴(𝑋 × 𝑋)𝐵 ∧ ¬ 𝐴( < < )𝐵))
75, 6bitri 275 . . . 4 (𝐴𝑅𝐵 ↔ (𝐴(𝑋 × 𝑋)𝐵 ∧ ¬ 𝐴( < < )𝐵))
87baib 535 . . 3 (𝐴(𝑋 × 𝑋)𝐵 → (𝐴𝑅𝐵 ↔ ¬ 𝐴( < < )𝐵))
93, 8syl 17 . 2 ((𝐴𝑋𝐵𝑋) → (𝐴𝑅𝐵 ↔ ¬ 𝐴( < < )𝐵))
10 brun 5194 . . . 4 (𝐴( < < )𝐵 ↔ (𝐴 < 𝐵𝐴 < 𝐵))
11 brcnvg 5890 . . . . 5 ((𝐴𝑋𝐵𝑋) → (𝐴 < 𝐵𝐵 < 𝐴))
1211orbi2d 916 . . . 4 ((𝐴𝑋𝐵𝑋) → ((𝐴 < 𝐵𝐴 < 𝐵) ↔ (𝐴 < 𝐵𝐵 < 𝐴)))
1310, 12bitrid 283 . . 3 ((𝐴𝑋𝐵𝑋) → (𝐴( < < )𝐵 ↔ (𝐴 < 𝐵𝐵 < 𝐴)))
1413notbid 318 . 2 ((𝐴𝑋𝐵𝑋) → (¬ 𝐴( < < )𝐵 ↔ ¬ (𝐴 < 𝐵𝐵 < 𝐴)))
159, 14bitrd 279 1 ((𝐴𝑋𝐵𝑋) → (𝐴𝑅𝐵 ↔ ¬ (𝐴 < 𝐵𝐵 < 𝐴)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 206  wa 395  wo 848   = wceq 1540  wcel 2108  cdif 3948  cun 3949  cop 4632   class class class wbr 5143   × cxp 5683  ccnv 5684
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pr 5432
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-br 5144  df-opab 5206  df-xp 5691  df-cnv 5693
This theorem is referenced by:  swoer  8776  swoord1  8777  swoord2  8778  swoso  8779
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