Theorem unss2 4150

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 4148	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))
2		uncom 4121	. 2 ⊢ (𝐶 ∪ 𝐴) = (𝐴 ∪ 𝐶)
3		uncom 4121	. 2 ⊢ (𝐶 ∪ 𝐵) = (𝐵 ∪ 𝐶)
4	1, 2, 3	3sstr4g 4000	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∪ 𝐴) ⊆ (𝐶 ∪ 𝐵))

Syntax hints:   → wi 4   ∪ cun 3912   ⊆ wss 3914

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 2008  ax-8 2111  ax-9 2119  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-v 3449  df-un 3919  df-ss 3931

This theorem is referenced by:  unss12  4151  ord3ex  5342  xpider  8761  fin1a2lem13  10365  canthp1lem2  10606  seqexw  13982  uniioombllem3  25486  volcn  25507  dvres2lem  25811  mulsproplem13  28031  mulsproplem14  28032  bnj1413  35025  bnj1408  35026