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Theorem ssdisj 4412
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 4194 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 3992 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3947 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4354 . 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 3900  wss 3901  c0 4285
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 2115  ax-9 2123  ax-ext 2708
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 2715  df-cleq 2728  df-clel 2811  df-v 3442  df-dif 3904  df-in 3908  df-ss 3918  df-nul 4286
This theorem is referenced by:  djudisj  6125  fimacnvdisj  6712  marypha1lem  9336  djuin  9830  ackbij1lem16  10144  ackbij1lem18  10146  fin23lem20  10247  fin23lem30  10252  psdmul  22109  elcls3  23027  neindisj  23061  imadifxp  32676  ldgenpisyslem1  34320  chtvalz  34786  pthhashvtx  35322  diophren  43055
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