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Theorem inss 4144
Description: Inclusion of an intersection of two classes. (Contributed by NM, 30-Oct-2014.)
Assertion
Ref Expression
inss ((𝐴 ⊆ 𝐶 ∨ 𝐵 ⊆ 𝐶) → (𝐴 ∩ 𝐵) ⊆ 𝐶)

Proof of Theorem inss
StepHypRef Expression
1 ssinss1 4141 . 2 (𝐴 ⊆ 𝐶 → (𝐴 ∩ 𝐵) ⊆ 𝐶)
2 incom 4105 . . 3 (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
3 ssinss1 4141 . . 3 (𝐵 ⊆ 𝐶 → (𝐵 ∩ 𝐴) ⊆ 𝐶)
42, 3eqsstrid 3947 . 2 (𝐵 ⊆ 𝐶 → (𝐴 ∩ 𝐵) ⊆ 𝐶)
51, 4jaoi 871 1 ((𝐴 ⊆ 𝐶 ∨ 𝐵 ⊆ 𝐶) → (𝐴 ∩ 𝐵) ⊆ 𝐶)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ∨ wo 861   ⊆ wss 3896   ∩ cin 4088
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3897  df-in 4090
This theorem is used by:  pmatcoe1fsupp  23028  ppttop  23334  disjorimxrn  39917  iunrelexp0  44839  ntrclsk3  45207  icccncfext  47033
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