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Theorem undisj1 3317
 Description: The union of disjoint classes is disjoint. (Contributed by NM, 26-Sep-2004.)
Assertion
Ref Expression
undisj1 (((𝐴𝐶) = ∅ ∧ (𝐵𝐶) = ∅) ↔ ((𝐴𝐵) ∩ 𝐶) = ∅)

Proof of Theorem undisj1
StepHypRef Expression
1 un00 3306 . 2 (((𝐴𝐶) = ∅ ∧ (𝐵𝐶) = ∅) ↔ ((𝐴𝐶) ∪ (𝐵𝐶)) = ∅)
2 indir 3229 . . 3 ((𝐴𝐵) ∩ 𝐶) = ((𝐴𝐶) ∪ (𝐵𝐶))
32eqeq1i 2090 . 2 (((𝐴𝐵) ∩ 𝐶) = ∅ ↔ ((𝐴𝐶) ∪ (𝐵𝐶)) = ∅)
41, 3bitr4i 185 1 (((𝐴𝐶) = ∅ ∧ (𝐵𝐶) = ∅) ↔ ((𝐴𝐵) ∩ 𝐶) = ∅)
 Colors of variables: wff set class Syntax hints:   ∧ wa 102   ↔ wb 103   = wceq 1285   ∪ cun 2980   ∩ cin 2981  ∅c0 3267 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065 This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-v 2612  df-dif 2984  df-un 2986  df-in 2988  df-ss 2995  df-nul 3268 This theorem is referenced by:  funtp  5003
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