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Theorem inss 4201
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 4198 . 2 (𝐴𝐶 → (𝐴𝐵) ⊆ 𝐶)
2 incom 4162 . . 3 (𝐴𝐵) = (𝐵𝐴)
3 ssinss1 4198 . . 3 (𝐵𝐶 → (𝐵𝐴) ⊆ 𝐶)
42, 3eqsstrid 3975 . 2 (𝐵𝐶 → (𝐴𝐵) ⊆ 𝐶)
51, 4jaoi 870 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵) ⊆ 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wo 860  cin 3904  wss 3905
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-in 3912  df-ss 3922
This theorem is referenced by:  pmatcoe1fsupp  22858  ppttop  23164  disjorimxrn  39497  iunrelexp0  44428  ntrclsk3  44796  icccncfext  46601
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