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Theorem ssdifin0 4451
Description: A subset of a difference does not intersect the subtrahend. (Contributed by Jeff Hankins, 1-Sep-2013.) (Proof shortened by Mario Carneiro, 24-Aug-2015.)
Assertion
Ref Expression
ssdifin0 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) = ∅)

Proof of Theorem ssdifin0
StepHypRef Expression
1 ssrin 4202 . 2 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶))
2 disjdifr 4439 . 2 ((𝐵𝐶) ∩ 𝐶) = ∅
3 sseq0 4367 . 2 (((𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶) ∧ ((𝐵𝐶) ∩ 𝐶) = ∅) → (𝐴𝐶) = ∅)
41, 2, 3sylancl 597 1 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1567  cdif 3910  cin 3912  wss 3913  c0 4294
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-rab 3424  df-v 3465  df-dif 3916  df-in 3920  df-ss 3930  df-nul 4295
This theorem is referenced by:  ssdifeq0  4452  marypha1lem  9393  numacn  10033  mreexexlem2d  17701  mreexexlem4d  17703  nrmsep2  23482  isnrm3  23485
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