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Theorem ssdisj 4407
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 4208 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 4021 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3978 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4350 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
53, 4syl 17 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 398   = wceq 1531  cin 3933  wss 3934  c0 4289
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1905  ax-6 1964  ax-7 2009  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2154  ax-12 2170  ax-ext 2791
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1534  df-ex 1775  df-nf 1779  df-sb 2064  df-clab 2798  df-cleq 2812  df-clel 2891  df-nfc 2961  df-v 3495  df-dif 3937  df-in 3941  df-ss 3950  df-nul 4290
This theorem is referenced by:  djudisj  6017  fimacnvdisj  6550  marypha1lem  8889  djuin  9339  ackbij1lem16  9649  ackbij1lem18  9651  fin23lem20  9751  fin23lem30  9756  elcls3  21683  neindisj  21717  imadifxp  30343  ldgenpisyslem1  31415  chtvalz  31893  pthhashvtx  32367  diophren  39400
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