Theorem unss1 4131

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 3925	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	orim1d 981	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
3		elun 4100	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))
4		elun 4100	. . 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 3937	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))

This proof depends on syntax axioms:   → wi 4   ∨ wo 861   ∈ wcel 2145   ∪ cun 3897   ⊆ wss 3899

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-un 3904  df-ss 3916

This theorem is used by:  unss2  4133  unss12  4134  eldifpw  7767  orderseqlem  8155  tposss  8225  dftpos4  8243  hashbclem  14517  incexclem  15925  mreexexlem2d  17733  catcoppccl  18206  neitr  23405  restntr  23407  leordtval2  23437  cmpcld  23627  uniioombllem3  25813  limcres  26113  plyss  26424  mulsproplem13  28393  mulsproplem14  28394  shlej1  31841  fineqvac  35642  ss2mcls  36147  bj-rrhatsscchat  37988  pclfinclN  40823  dmtposss  49802