MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  ssdisj Structured version   Visualization version   GIF version

Theorem ssdisj 4423
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 4205 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 4005 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3960 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4365 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
53, 4syl 17 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1540  cin 3913  wss 3914  c0 4296
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 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-v 3449  df-dif 3917  df-in 3921  df-ss 3931  df-nul 4297
This theorem is referenced by:  djudisj  6140  fimacnvdisj  6738  marypha1lem  9384  djuin  9871  ackbij1lem16  10187  ackbij1lem18  10189  fin23lem20  10290  fin23lem30  10295  psdmul  22053  elcls3  22970  neindisj  23004  imadifxp  32530  ldgenpisyslem1  34153  chtvalz  34620  pthhashvtx  35115  diophren  42801
  Copyright terms: Public domain W3C validator