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Theorem brdif 3915
 Description: The difference of two binary relations. (Contributed by Scott Fenton, 11-Apr-2011.)
Assertion
Ref Expression
brdif (𝐴(𝑅𝑆)𝐵 ↔ (𝐴𝑅𝐵 ∧ ¬ 𝐴𝑆𝐵))

Proof of Theorem brdif
StepHypRef Expression
1 eldif 3022 . 2 (⟨𝐴, 𝐵⟩ ∈ (𝑅𝑆) ↔ (⟨𝐴, 𝐵⟩ ∈ 𝑅 ∧ ¬ ⟨𝐴, 𝐵⟩ ∈ 𝑆))
2 df-br 3868 . 2 (𝐴(𝑅𝑆)𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑅𝑆))
3 df-br 3868 . . 3 (𝐴𝑅𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑅)
4 df-br 3868 . . . 4 (𝐴𝑆𝐵 ↔ ⟨𝐴, 𝐵⟩ ∈ 𝑆)
54notbii 632 . . 3 𝐴𝑆𝐵 ↔ ¬ ⟨𝐴, 𝐵⟩ ∈ 𝑆)
63, 5anbi12i 449 . 2 ((𝐴𝑅𝐵 ∧ ¬ 𝐴𝑆𝐵) ↔ (⟨𝐴, 𝐵⟩ ∈ 𝑅 ∧ ¬ ⟨𝐴, 𝐵⟩ ∈ 𝑆))
71, 2, 63bitr4i 211 1 (𝐴(𝑅𝑆)𝐵 ↔ (𝐴𝑅𝐵 ∧ ¬ 𝐴𝑆𝐵))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   ∧ wa 103   ↔ wb 104   ∈ wcel 1445   ∖ cdif 3010  ⟨cop 3469   class class class wbr 3867 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 582  ax-in2 583  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077 This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-v 2635  df-dif 3015  df-br 3868 This theorem is referenced by:  fndmdif  5443  brdifun  6359
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