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Theorem ssdisj 4410
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 4192 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 3993 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3948 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4352 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
53, 4syl 17 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1541  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 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-v 3438  df-dif 3905  df-in 3909  df-ss 3919  df-nul 4284
This theorem is referenced by:  djudisj  6114  fimacnvdisj  6701  marypha1lem  9317  djuin  9811  ackbij1lem16  10125  ackbij1lem18  10127  fin23lem20  10228  fin23lem30  10233  psdmul  22082  elcls3  22999  neindisj  23033  imadifxp  32579  ldgenpisyslem1  34174  chtvalz  34640  pthhashvtx  35170  diophren  42852
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