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Theorem inss 4189
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 4187 . 2 (𝐴𝐶 → (𝐴𝐵) ⊆ 𝐶)
2 incom 4150 . . 3 (𝐴𝐵) = (𝐵𝐴)
3 ssinss1 4187 . . 3 (𝐵𝐶 → (𝐵𝐴) ⊆ 𝐶)
42, 3eqsstrid 3961 . 2 (𝐵𝐶 → (𝐴𝐵) ⊆ 𝐶)
51, 4jaoi 858 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵) ⊆ 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wo 848  cin 3889  wss 3890
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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3391  df-v 3432  df-in 3897  df-ss 3907
This theorem is referenced by:  pmatcoe1fsupp  22676  ppttop  22982  disjorimxrn  39183  iunrelexp0  44147  ntrclsk3  44515  icccncfext  46333
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