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Theorem ssdifin0 4429
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 4179 . 2 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶))
2 disjdifr 4418 . 2 ((𝐵𝐶) ∩ 𝐶) = ∅
3 sseq0 4345 . 2 (((𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶) ∧ ((𝐵𝐶) ∩ 𝐶) = ∅) → (𝐴𝐶) = ∅)
41, 2, 3sylancl 586 1 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) = ∅)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1540  cdif 3894  cin 3896  wss 3897  c0 4268
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2707
This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1543  df-fal 1553  df-ex 1781  df-sb 2067  df-clab 2714  df-cleq 2728  df-clel 2814  df-rab 3404  df-v 3443  df-dif 3900  df-in 3904  df-ss 3914  df-nul 4269
This theorem is referenced by:  ssdifeq0  4430  marypha1lem  9282  numacn  9898  mreexexlem2d  17443  mreexexlem4d  17445  nrmsep2  22605  isnrm3  22608
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