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Theorem ssdisj 4411
Description: Intersection with a subclass of a disjoint class. (Contributed by FL, 24-Jan-2007.) (Proof shortened by JJ, 14-Jul-2021.)
Assertion
Ref Expression
ssdisj ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)

Proof of Theorem ssdisj
StepHypRef Expression
1 ssrin 4191 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 3992 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3947 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4353 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
53, 4syl 17 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 399   = wceq 1559  cin 3901  wss 3902  c0 4283
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733
This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1562  df-fal 1572  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3455  df-dif 3905  df-in 3909  df-ss 3919  df-nul 4284
This theorem is referenced by:  djudisj  6147  fimacnvdisj  6736  marypha1lem  9372  djuin  9869  ackbij1lem16  10183  ackbij1lem18  10185  fin23lem20  10287  fin23lem30  10292  psdmul  22218  elcls3  23130  neindisj  23164  imadifxp  32760  ldgenpisyslem1  34420  chtvalz  34883  pthhashvtx  35438  diophren  43350
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