Theorem inss 4197

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 4195	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∩ 𝐵) ⊆ 𝐶)
2		incom 4158	. . 3 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
3		ssinss1 4195	. . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐵 ∩ 𝐴) ⊆ 𝐶)
4	2, 3	eqsstrid 3969	. 2 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∩ 𝐵) ⊆ 𝐶)
5	1, 4	jaoi 857	1 ⊢ ((𝐴 ⊆ 𝐶 ∨ 𝐵 ⊆ 𝐶) → (𝐴 ∩ 𝐵) ⊆ 𝐶)

Syntax hints:   → wi 4   ∨ wo 847   ∩ cin 3897   ⊆ wss 3898

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-rab 3397  df-v 3439  df-in 3905  df-ss 3915

This theorem is referenced by:  pmatcoe1fsupp  22636  ppttop  22942  disjorimxrn  38919  iunrelexp0  43859  ntrclsk3  44227  icccncfext  46047