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Theorem inss 4165
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 4164 . 2 (𝐴𝐶 → (𝐴𝐵) ⊆ 𝐶)
2 incom 4128 . . 3 (𝐴𝐵) = (𝐵𝐴)
3 ssinss1 4164 . . 3 (𝐵𝐶 → (𝐵𝐴) ⊆ 𝐶)
42, 3eqsstrid 3963 . 2 (𝐵𝐶 → (𝐴𝐵) ⊆ 𝐶)
51, 4jaoi 854 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵) ⊆ 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wo 844  cin 3880  wss 3881
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-rab 3115  df-v 3443  df-in 3888  df-ss 3898
This theorem is referenced by:  pmatcoe1fsupp  21306  ppttop  21612  inindif  30287  disjorimxrn  36138  iunrelexp0  40403  ntrclsk3  40773  icccncfext  42529
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