MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  inss Structured version   Visualization version   GIF version

Theorem inss 4193
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 4191 . 2 (𝐴𝐶 → (𝐴𝐵) ⊆ 𝐶)
2 incom 4154 . . 3 (𝐴𝐵) = (𝐵𝐴)
3 ssinss1 4191 . . 3 (𝐵𝐶 → (𝐵𝐴) ⊆ 𝐶)
42, 3eqsstrid 3968 . 2 (𝐵𝐶 → (𝐴𝐵) ⊆ 𝐶)
51, 4jaoi 857 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵) ⊆ 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wo 847  cin 3896  wss 3897
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 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-v 3438  df-in 3904  df-ss 3914
This theorem is referenced by:  pmatcoe1fsupp  22611  ppttop  22917  disjorimxrn  38786  iunrelexp0  43735  ntrclsk3  44103  icccncfext  45925
  Copyright terms: Public domain W3C validator