Theorem inss 4148

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

Step	Hyp	Ref	Expression
1		ssinss1 4145	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∩ 𝐵) ⊆ 𝐶)
2		incom 4109	. . 3 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
3		ssinss1 4145	. . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐵 ∩ 𝐴) ⊆ 𝐶)
4	2, 3	eqsstrid 3951	. 2 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∩ 𝐵) ⊆ 𝐶)
5	1, 4	jaoi 871	1 ⊢ ((𝐴 ⊆ 𝐶 ∨ 𝐵 ⊆ 𝐶) → (𝐴 ∩ 𝐵) ⊆ 𝐶)

This proof depends on syntax axioms:   → wi 4   ∨ wo 861   ⊆ wss 3900   ∩ cin 4092

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-in 4094

This theorem is used by:  pmatcoe1fsupp  23074  ppttop  23380  disjorimxrn  40046  iunrelexp0  44963  ntrclsk3  45331  icccncfext  47170