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Theorem ssdisj 3477
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 3460 . . . 4 ((𝐵𝐶) ⊆ ∅ ↔ (𝐵𝐶) = ∅)
2 ssrin 3358 . . . . 5 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
3 sstr2 3160 . . . . 5 ((𝐴𝐶) ⊆ (𝐵𝐶) → ((𝐵𝐶) ⊆ ∅ → (𝐴𝐶) ⊆ ∅))
42, 3syl 14 . . . 4 (𝐴𝐵 → ((𝐵𝐶) ⊆ ∅ → (𝐴𝐶) ⊆ ∅))
51, 4syl5bir 153 . . 3 (𝐴𝐵 → ((𝐵𝐶) = ∅ → (𝐴𝐶) ⊆ ∅))
65imp 124 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
7 ss0 3461 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
86, 7syl 14 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104   = wceq 1353  cin 3126  wss 3127  c0 3420
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615  ax-io 709  ax-5 1445  ax-7 1446  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-8 1502  ax-10 1503  ax-11 1504  ax-i12 1505  ax-bndl 1507  ax-4 1508  ax-17 1524  ax-i9 1528  ax-ial 1532  ax-i5r 1533  ax-ext 2157
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1459  df-sb 1761  df-clab 2162  df-cleq 2168  df-clel 2171  df-nfc 2306  df-v 2737  df-dif 3129  df-in 3133  df-ss 3140  df-nul 3421
This theorem is referenced by:  djudisj  5048  fimacnvdisj  5392  unfiin  6915  hashunlem  10752
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