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Theorem ssdisj 3389
Description: Intersection with a subclass of a disjoint class. (Contributed by FL, 24-Jan-2007.)
Assertion
Ref Expression
ssdisj ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)

Proof of Theorem ssdisj
StepHypRef Expression
1 ss0b 3372 . . . 4 ((𝐵𝐶) ⊆ ∅ ↔ (𝐵𝐶) = ∅)
2 ssrin 3271 . . . . 5 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
3 sstr2 3074 . . . . 5 ((𝐴𝐶) ⊆ (𝐵𝐶) → ((𝐵𝐶) ⊆ ∅ → (𝐴𝐶) ⊆ ∅))
42, 3syl 14 . . . 4 (𝐴𝐵 → ((𝐵𝐶) ⊆ ∅ → (𝐴𝐶) ⊆ ∅))
51, 4syl5bir 152 . . 3 (𝐴𝐵 → ((𝐵𝐶) = ∅ → (𝐴𝐶) ⊆ ∅))
65imp 123 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
7 ss0 3373 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
86, 7syl 14 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103   = wceq 1316  cin 3040  wss 3041  c0 3333
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 588  ax-in2 589  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-v 2662  df-dif 3043  df-in 3047  df-ss 3054  df-nul 3334
This theorem is referenced by:  djudisj  4936  fimacnvdisj  5277  unfiin  6782  hashunlem  10518
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