Theorem unss1 3433

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 3268	. . . 4 ⊢ (A ⊆ B → (x ∈ A → x ∈ B))
2	1	orim1d 812	. . 3 ⊢ (A ⊆ B → ((x ∈ A ∨ x ∈ C) → (x ∈ B ∨ x ∈ C)))
3		elun 3221	. . 3 ⊢ (x ∈ (A ∪ C) ↔ (x ∈ A ∨ x ∈ C))
4		elun 3221	. . 3 ⊢ (x ∈ (B ∪ C) ↔ (x ∈ B ∨ x ∈ C))
5	2, 3, 4	3imtr4g 261	. 2 ⊢ (A ⊆ B → (x ∈ (A ∪ C) → x ∈ (B ∪ C)))
6	5	ssrdv 3279	1 ⊢ (A ⊆ B → (A ∪ C) ⊆ (B ∪ C))

This proof depends on syntax axioms:   → wi 4   ∨ wo 357   ∈ wcel 1710   ∪ cun 3208   ⊆ wss 3258

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-ss 3260

This theorem is used by:  unss2  3435  unss12  3436  pwadjoin  4120