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Theorem ssdifin0 4440
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 4196 . 2 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶))
2 disjdifr 4427 . 2 ((𝐵𝐶) ∩ 𝐶) = ∅
3 sseq0 4357 . 2 (((𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶) ∧ ((𝐵𝐶) ∩ 𝐶) = ∅) → (𝐴𝐶) = ∅)
41, 2, 3sylancl 587 1 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1542  cdif 3900  cin 3902  wss 3903  c0 4287
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3402  df-v 3444  df-dif 3906  df-in 3910  df-ss 3920  df-nul 4288
This theorem is referenced by:  ssdifeq0  4441  marypha1lem  9350  numacn  9973  mreexexlem2d  17582  mreexexlem4d  17584  nrmsep2  23317  isnrm3  23320
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