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Theorem inss 4169
Description: Inclusion of an intersection of two classes. (Contributed by NM, 30-Oct-2014.)
Assertion
Ref Expression
inss ((𝐴𝐶𝐵𝐶) → (𝐴𝐵) ⊆ 𝐶)

Proof of Theorem inss
StepHypRef Expression
1 ssinss1 4168 . 2 (𝐴𝐶 → (𝐴𝐵) ⊆ 𝐶)
2 incom 4131 . . 3 (𝐴𝐵) = (𝐵𝐴)
3 ssinss1 4168 . . 3 (𝐵𝐶 → (𝐵𝐴) ⊆ 𝐶)
42, 3eqsstrid 3965 . 2 (𝐵𝐶 → (𝐴𝐵) ⊆ 𝐶)
51, 4jaoi 853 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵) ⊆ 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wo 843  cin 3882  wss 3883
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-rab 3072  df-v 3424  df-in 3890  df-ss 3900
This theorem is referenced by:  pmatcoe1fsupp  21758  ppttop  22065  inindif  30764  disjorimxrn  36783  iunrelexp0  41199  ntrclsk3  41569  icccncfext  43318
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