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Theorem ssdisj 3316
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 3299 . . . 4 ((𝐵𝐶) ⊆ ∅ ↔ (𝐵𝐶) = ∅)
2 ssrin 3207 . . . . 5 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
3 sstr2 3015 . . . . 5 ((𝐴𝐶) ⊆ (𝐵𝐶) → ((𝐵𝐶) ⊆ ∅ → (𝐴𝐶) ⊆ ∅))
42, 3syl 14 . . . 4 (𝐴𝐵 → ((𝐵𝐶) ⊆ ∅ → (𝐴𝐶) ⊆ ∅))
51, 4syl5bir 151 . . 3 (𝐴𝐵 → ((𝐵𝐶) = ∅ → (𝐴𝐶) ⊆ ∅))
65imp 122 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
7 ss0 3300 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
86, 7syl 14 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102   = wceq 1285  cin 2981  wss 2982  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-in 2988  df-ss 2995  df-nul 3268
This theorem is referenced by:  djudisj  4800  fimacnvdisj  5125  unfiin  6471  djuin  6562  hashunlem  9898
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