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Theorem ssdisj 4417
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 4196 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 3997 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3952 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4359 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
53, 4syl 18 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 400   = wceq 1563  cin 3906  wss 3907  c0 4288
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-ext 2737
This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1566  df-fal 1576  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-dif 3910  df-in 3914  df-ss 3924  df-nul 4289
This theorem is referenced by:  djudisj  6155  fimacnvdisj  6746  marypha1lem  9381  djuin  9892  ackbij1lem16  10205  ackbij1lem18  10207  fin23lem20  10309  fin23lem30  10314  psdmul  22286  elcls3  23197  neindisj  23231  imadifxp  32852  ldgenpisyslem1  34465  chtvalz  34928  pthhashvtx  35486  diophren  43397
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