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Theorem ssdisj 3480
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 3463 . . . 4 ((𝐵𝐶) ⊆ ∅ ↔ (𝐵𝐶) = ∅)
2 ssrin 3361 . . . . 5 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
3 sstr2 3163 . . . . 5 ((𝐴𝐶) ⊆ (𝐵𝐶) → ((𝐵𝐶) ⊆ ∅ → (𝐴𝐶) ⊆ ∅))
42, 3syl 14 . . . 4 (𝐴𝐵 → ((𝐵𝐶) ⊆ ∅ → (𝐴𝐶) ⊆ ∅))
51, 4biimtrrid 153 . . 3 (𝐴𝐵 → ((𝐵𝐶) = ∅ → (𝐴𝐶) ⊆ ∅))
65imp 124 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
7 ss0 3464 . 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 3129  wss 3130  c0 3423
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 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2740  df-dif 3132  df-in 3136  df-ss 3143  df-nul 3424
This theorem is referenced by:  djudisj  5057  fimacnvdisj  5401  unfiin  6925  hashunlem  10784
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