Theorem unss1 4183

Description: Subclass law for union of classes. (Contributed by NM, 14-Oct-1999.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
unss1	⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))

Proof of Theorem unss1

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ssel 3902	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	orim1d 981	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
3		elun 4152	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))
4		elun 4152	. . 3 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))
5	2, 3, 4	3imtr4g 299	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ (𝐴 ∪ 𝐶) → 𝑥 ∈ (𝐵 ∪ 𝐶)))
6	5	ssrdv 3914	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))

This proof depends on syntax axioms:   → wi 4   ∨ wo 861   ∈ wcel 2144   ⊆ wss 3895   ∪ cun 4149

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-ss 3896  df-un 4151

This theorem is used by:  unss2  4185  unss12  4186  eldifpw  7782  orderseqlem  8174  tposss  8244  dftpos4  8262  hashbclem  14597  incexclem  16005  mreexexlem2d  17819  catcoppccl  18292  neitr  23498  restntr  23500  leordtval2  23530  cmpcld  23720  uniioombllem3  25906  limcres  26206  plyss  26517  mulsproplem13  28514  mulsproplem14  28515  shlej1  31962  fineqvac  35784  ss2mcls  36333  bj-rrhatsscchat  38158  pclfinclN  41008  dmtposss  49984