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Theorem brdifun 8665
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 5656 . . . 4 ((𝐴𝑋𝐵𝑋) → ⟨𝐴, 𝐵⟩ ∈ (𝑋 × 𝑋))
2 df-br 5074 . . . 4 (𝐴(𝑋 × 𝑋)𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑋 × 𝑋))
31, 2sylibr 235 . . 3 ((𝐴𝑋𝐵𝑋) → 𝐴(𝑋 × 𝑋)𝐵)
4 swoer.1 . . . . . 6 𝑅 = ((𝑋 × 𝑋) ∖ ( < < ))
54breqi 5079 . . . . 5 (𝐴𝑅𝐵𝐴((𝑋 × 𝑋) ∖ ( < < ))𝐵)
6 brdif 5126 . . . . 5 (𝐴((𝑋 × 𝑋) ∖ ( < < ))𝐵 ↔ (𝐴(𝑋 × 𝑋)𝐵 ∧ ¬ 𝐴( < < )𝐵))
75, 6bitri 276 . . . 4 (𝐴𝑅𝐵 ↔ (𝐴(𝑋 × 𝑋)𝐵 ∧ ¬ 𝐴( < < )𝐵))
87baib 540 . . 3 (𝐴(𝑋 × 𝑋)𝐵 → (𝐴𝑅𝐵 ↔ ¬ 𝐴( < < )𝐵))
93, 8syl 17 . 2 ((𝐴𝑋𝐵𝑋) → (𝐴𝑅𝐵 ↔ ¬ 𝐴( < < )𝐵))
10 brun 5124 . . . 4 (𝐴( < < )𝐵 ↔ (𝐴 < 𝐵𝐴 < 𝐵))
11 brcnvg 5822 . . . . 5 ((𝐴𝑋𝐵𝑋) → (𝐴 < 𝐵𝐵 < 𝐴))
1211orbi2d 921 . . . 4 ((𝐴𝑋𝐵𝑋) → ((𝐴 < 𝐵𝐴 < 𝐵) ↔ (𝐴 < 𝐵𝐵 < 𝐴)))
1310, 12bitrid 284 . . 3 ((𝐴𝑋𝐵𝑋) → (𝐴( < < )𝐵 ↔ (𝐴 < 𝐵𝐵 < 𝐴)))
1413notbid 319 . 2 ((𝐴𝑋𝐵𝑋) → (¬ 𝐴( < < )𝐵 ↔ ¬ (𝐴 < 𝐵𝐵 < 𝐴)))
159, 14bitrd 280 1 ((𝐴𝑋𝐵𝑋) → (𝐴𝑅𝐵 ↔ ¬ (𝐴 < 𝐵𝐵 < 𝐴)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 207  wa 396  wo 853   = wceq 1547  wcel 2119  cdif 3880  cun 3881  cop 4562   class class class wbr 5073   × cxp 5617  ccnv 5618
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711  ax-sep 5219  ax-pr 5363
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-ral 3054  df-rex 3064  df-rab 3392  df-v 3433  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4263  df-if 4456  df-sn 4557  df-pr 4559  df-op 4563  df-br 5074  df-opab 5136  df-xp 5625  df-cnv 5627
This theorem is referenced by:  swoer  8666  swoord1  8667  swoord2  8668  swoso  8669
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