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Theorem ssdifin0 4441
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 4195 . 2 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶))
2 disjdifr 4429 . 2 ((𝐵𝐶) ∩ 𝐶) = ∅
3 sseq0 4359 . 2 (((𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶) ∧ ((𝐵𝐶) ∩ 𝐶) = ∅) → (𝐴𝐶) = ∅)
41, 2, 3sylancl 595 1 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1562  cdif 3903  cin 3905  wss 3906  c0 4287
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-ext 2736
This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1565  df-fal 1575  df-ex 1802  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-rab 3417  df-v 3458  df-dif 3909  df-in 3913  df-ss 3923  df-nul 4288
This theorem is referenced by:  ssdifeq0  4442  marypha1lem  9381  numacn  10007  mreexexlem2d  17679  mreexexlem4d  17681  nrmsep2  23418  isnrm3  23421
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