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Theorem ssdisj 4459
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 4241 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 4041 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3996 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4401 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
53, 4syl 17 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1539  cin 3949  wss 3950  c0 4332
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2707
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1542  df-fal 1552  df-ex 1779  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-v 3481  df-dif 3953  df-in 3957  df-ss 3967  df-nul 4333
This theorem is referenced by:  djudisj  6186  fimacnvdisj  6785  marypha1lem  9474  djuin  9959  ackbij1lem16  10275  ackbij1lem18  10277  fin23lem20  10378  fin23lem30  10383  psdmul  22171  elcls3  23092  neindisj  23126  imadifxp  32615  ldgenpisyslem1  34165  chtvalz  34645  pthhashvtx  35134  diophren  42829
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