Theorem unss2 4162

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 4160	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))
2		uncom 4133	. 2 ⊢ (𝐶 ∪ 𝐴) = (𝐴 ∪ 𝐶)
3		uncom 4133	. 2 ⊢ (𝐶 ∪ 𝐵) = (𝐵 ∪ 𝐶)
4	1, 2, 3	3sstr4g 4012	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∪ 𝐴) ⊆ (𝐶 ∪ 𝐵))

Syntax hints:   → wi 4   ∪ cun 3924   ⊆ wss 3926

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-v 3461  df-un 3931  df-ss 3943

This theorem is referenced by:  unss12  4163  ord3ex  5357  xpider  8802  fin1a2lem13  10426  canthp1lem2  10667  seqexw  14035  uniioombllem3  25538  volcn  25559  dvres2lem  25863  mulsproplem13  28083  mulsproplem14  28084  bnj1413  35066  bnj1408  35067