Theorem unss1 3398

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 3242	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	orim1d 799	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶) → (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶)))
3		elun 3370	. . 3 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐶) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐶))
4		elun 3370	. . 3 ⊢ (𝑥 ∈ (𝐵 ∪ 𝐶) ↔ (𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐶))
5	2, 3, 4	3imtr4g 205	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ (𝐴 ∪ 𝐶) → 𝑥 ∈ (𝐵 ∪ 𝐶)))
6	5	ssrdv 3254	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∪ 𝐶) ⊆ (𝐵 ∪ 𝐶))

This proof depends on syntax axioms:   → wi 4   ∨ wo 720   ∈ wcel 2209   ∪ cun 3218   ⊆ wss 3220

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-un 3224  df-in 3226  df-ss 3233

This theorem is used by:  unss2  3400  unss12  3401  undif1ss  3602  eldifpw  4623  tposss  6517  dftpos4  6534  hashfibclem  11298  plyss  15930