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Theorem ssdifin0 4444
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 4190 . 2 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶))
2 disjdifr 4430 . 2 ((𝐵𝐶) ∩ 𝐶) = ∅
3 sseq0 4357 . 2 (((𝐴𝐶) ⊆ ((𝐵𝐶) ∩ 𝐶) ∧ ((𝐵𝐶) ∩ 𝐶) = ∅) → (𝐴𝐶) = ∅)
41, 2, 3sylancl 598 1 (𝐴 ⊆ (𝐵𝐶) → (𝐴𝐶) = ∅)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4   = wceq 1570  cdif 3899  cin 3901  wss 3902  c0 4282
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3455  df-dif 3905  df-in 3909  df-ss 3919  df-nul 4283
This theorem is used by:  ssdifeq0  4445  marypha1lem  9407  numacn  10056  mreexexlem2d  17739  mreexexlem4d  17741  nrmsep2  23587  isnrm3  23590
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