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Theorem ssdisj 4458
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 4232 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 4039 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3994 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4397 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
53, 4syl 17 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 394   = wceq 1539  cin 3946  wss 3947  c0 4321
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701
This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1542  df-fal 1552  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-v 3474  df-dif 3950  df-in 3954  df-ss 3964  df-nul 4322
This theorem is referenced by:  djudisj  6165  fimacnvdisj  6768  marypha1lem  9430  djuin  9915  ackbij1lem16  10232  ackbij1lem18  10234  fin23lem20  10334  fin23lem30  10339  elcls3  22807  neindisj  22841  imadifxp  32099  ldgenpisyslem1  33459  chtvalz  33939  pthhashvtx  34416  diophren  41853
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