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Theorem ssdifin0 4477
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 4225 . 2 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶))
2 disjdifr 4464 . 2 ((𝐵𝐶) ∩ 𝐶) = ∅
3 sseq0 4391 . 2 (((𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶) ∧ ((𝐵𝐶) ∩ 𝐶) = ∅) → (𝐴𝐶) = ∅)
41, 2, 3sylancl 585 1 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1533  cdif 3937  cin 3939  wss 3940  c0 4314
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2695
This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802  df-rab 3425  df-v 3468  df-dif 3943  df-in 3947  df-ss 3957  df-nul 4315
This theorem is referenced by:  ssdifeq0  4478  marypha1lem  9423  numacn  10039  mreexexlem2d  17585  mreexexlem4d  17587  nrmsep2  23170  isnrm3  23173
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