Theorem unss2 4111

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 4109	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))
2		uncom 4083	. 2 ⊢ (𝐶 ∪ 𝐴) = (𝐴 ∪ 𝐶)
3		uncom 4083	. 2 ⊢ (𝐶 ∪ 𝐵) = (𝐵 ∪ 𝐶)
4	1, 2, 3	3sstr4g 3962	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∪ 𝐴) ⊆ (𝐶 ∪ 𝐵))

Syntax hints:   → wi 4   ∪ cun 3881   ⊆ wss 3883

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3424  df-un 3888  df-in 3890  df-ss 3900

This theorem is referenced by:  unss12  4112  ord3ex  5305  xpider  8535  fin1a2lem13  10099  canthp1lem2  10340  seqexw  13665  uniioombllem3  24654  volcn  24675  dvres2lem  24979  bnj1413  32915  bnj1408  32916