Theorem unss2 4141

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 4139	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))
2		uncom 4113	. 2 ⊢ (𝐶 ∪ 𝐴) = (𝐴 ∪ 𝐶)
3		uncom 4113	. 2 ⊢ (𝐶 ∪ 𝐵) = (𝐵 ∪ 𝐶)
4	1, 2, 3	3sstr4g 3991	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∪ 𝐴) ⊆ (𝐶 ∪ 𝐵))

Syntax hints:   → wi 4   ∪ cun 3904   ⊆ wss 3906

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3457  df-un 3911  df-ss 3923

This theorem is referenced by:  unss12  4142  ord3ex  5360  xpider  8787  fin1a2lem13  10397  canthp1lem2  10639  seqexw  14055  uniioombllem3  25725  volcn  25746  dvres2lem  26050  mulsproplem13  28302  mulsproplem14  28303  bnj1413  35404  bnj1408  35405