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Theorem ssdisj 4395
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 4177 . . 3 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 eqimss 3980 . . 3 ((𝐵𝐶) = ∅ → (𝐵𝐶) ⊆ ∅)
31, 2sylan9ss 3935 . 2 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) ⊆ ∅)
4 ss0 4337 . 2 ((𝐴𝐶) ⊆ ∅ → (𝐴𝐶) = ∅)
53, 4syl 17 1 ((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 396   = wceq 1547  cin 3889  wss 3890  c0 4268
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712
This theorem depends on definitions:  df-bi 208  df-an 397  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-v 3434  df-dif 3893  df-in 3897  df-ss 3907  df-nul 4269
This theorem is referenced by:  djudisj  6125  fimacnvdisj  6712  marypha1lem  9343  djuin  9840  ackbij1lem16  10154  ackbij1lem18  10156  fin23lem20  10257  fin23lem30  10262  psdmul  22161  elcls3  23073  neindisj  23107  imadifxp  32697  ldgenpisyslem1  34354  chtvalz  34820  pthhashvtx  35363  diophren  43265
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