Theorem unss2 3160

Description: Subclass law for union of classes. Exercise 7 of [TakeutiZaring] p. 18. (Contributed by NM, 14-Oct-1999.)

Assertion

Ref	Expression
unss2	⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∪ 𝐴) ⊆ (𝐶 ∪ 𝐵))

Proof of Theorem unss2

Step	Hyp	Ref	Expression
1		unss1 3158	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))
2		uncom 3133	. 2 ⊢ (𝐶 ∪ 𝐴) = (𝐴 ∪ 𝐶)
3		uncom 3133	. 2 ⊢ (𝐶 ∪ 𝐵) = (𝐵 ∪ 𝐶)
4	1, 2, 3	3sstr4g 3056	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∪ 𝐴) ⊆ (𝐶 ∪ 𝐵))

Syntax hints:   → wi 4   ∪ cun 2986   ⊆ wss 2988

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1379  ax-7 1380  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-8 1438  ax-10 1439  ax-11 1440  ax-i12 1441  ax-bndl 1442  ax-4 1443  ax-17 1462  ax-i9 1466  ax-ial 1470  ax-i5r 1471  ax-ext 2067

This theorem depends on definitions:  df-bi 115  df-tru 1290  df-nf 1393  df-sb 1690  df-clab 2072  df-cleq 2078  df-clel 2081  df-nfc 2214  df-v 2617  df-un 2992  df-in 2994  df-ss 3001

This theorem is referenced by:  unss12  3161  difdif2ss  3245  difdifdirss  3354  ord3ex  4001  rdgss  6104  xpiderm  6317