Theorem unss1 4138

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 3932	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	orim1d 981	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
3		elun 4107	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))
4		elun 4107	. . 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 3944	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))

This proof depends on syntax axioms:   → wi 4   ∨ wo 861   ∈ wcel 2146   ∪ cun 3904   ⊆ wss 3906

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 2148  ax-9 2156  ax-ext 2737

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 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-un 3911  df-ss 3923

This theorem is used by:  unss2  4140  unss12  4141  eldifpw  7773  orderseqlem  8159  tposss  8229  dftpos4  8247  hashbclem  14507  incexclem  15913  mreexexlem2d  17723  catcoppccl  18196  neitr  23387  restntr  23389  leordtval2  23419  cmpcld  23609  uniioombllem3  25795  limcres  26096  plyss  26407  mulsproplem13  28372  mulsproplem14  28373  shlej1  31783  fineqvac  35586  ss2mcls  36097  bj-rrhatsscchat  37937  pclfinclN  40782  dmtposss  49711