Theorem unss1 4186

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 3907	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	orim1d 981	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
3		elun 4155	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))
4		elun 4155	. . 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 3919	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))

This proof depends on syntax axioms:   → wi 4   ∨ wo 861   ∈ wcel 2147   ⊆ wss 3900   ∪ cun 4152

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-ss 3901  df-un 4154

This theorem is used by:  unss2  4188  unss12  4189  eldifpw  7814  orderseqlem  8211  tposss  8282  dftpos4  8300  hashbclem  14638  incexclem  16049  mreexexlem2d  17869  catcoppccl  18342  neitr  23553  restntr  23555  leordtval2  23585  cmpcld  23775  uniioombllem3  25961  limcres  26261  plyss  26572  mulsproplem13  28571  mulsproplem14  28572  shlej1  32019  fineqvac  35925  ss2mcls  36541  bj-rrhatsscchat  38425  pclfinclN  41273  dmtposss  50320