Theorem inss 3389

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 3388	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∩ 𝐵) ⊆ 𝐶)
2		incom 3351	. . 3 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
3		ssinss1 3388	. . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐵 ∩ 𝐴) ⊆ 𝐶)
4	2, 3	eqsstrid 3225	. 2 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∩ 𝐵) ⊆ 𝐶)
5	1, 4	jaoi 717	1 ⊢ ((𝐴 ⊆ 𝐶 ∨ 𝐵 ⊆ 𝐶) → (𝐴 ∩ 𝐵) ⊆ 𝐶)

Syntax hints:   → wi 4   ∨ wo 709   ∩ cin 3152   ⊆ wss 3153

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-in 3159  df-ss 3166

This theorem is referenced by: (None)