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Theorem inss 4200
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 4197 . 2 (𝐴𝐶 → (𝐴𝐵) ⊆ 𝐶)
2 incom 4161 . . 3 (𝐴𝐵) = (𝐵𝐴)
3 ssinss1 4197 . . 3 (𝐵𝐶 → (𝐵𝐴) ⊆ 𝐶)
42, 3eqsstrid 3974 . 2 (𝐵𝐶 → (𝐴𝐵) ⊆ 𝐶)
51, 4jaoi 868 1 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵) ⊆ 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wo 858  cin 3903  wss 3904
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734
This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1563  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3456  df-in 3911  df-ss 3921
This theorem is referenced by:  pmatcoe1fsupp  22758  ppttop  23064  disjorimxrn  39344  iunrelexp0  44275  ntrclsk3  44643  icccncfext  46458
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