Theorem rabun2 3535

Description: Abstraction restricted to a union. (Contributed by Stefan O'Rear, 5-Feb-2015.)

Assertion

Ref	Expression
rabun2	⊢ {x ∈ (A ∪ B) ∣ φ} = ({x ∈ A ∣ φ} ∪ {x ∈ B ∣ φ})

Proof of Theorem rabun2

Step	Hyp	Ref	Expression
1		df-rab 2624	. 2 ⊢ {x ∈ (A ∪ B) ∣ φ} = {x ∣ (x ∈ (A ∪ B) ∧ φ)}
2		df-rab 2624	. . . 4 ⊢ {x ∈ A ∣ φ} = {x ∣ (x ∈ A ∧ φ)}
3		df-rab 2624	. . . 4 ⊢ {x ∈ B ∣ φ} = {x ∣ (x ∈ B ∧ φ)}
4	2, 3	uneq12i 3417	. . 3 ⊢ ({x ∈ A ∣ φ} ∪ {x ∈ B ∣ φ}) = ({x ∣ (x ∈ A ∧ φ)} ∪ {x ∣ (x ∈ B ∧ φ)})
5		elun 3221	. . . . . . 7 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B))
6	5	anbi1i 676	. . . . . 6 ⊢ ((x ∈ (A ∪ B) ∧ φ) ↔ ((x ∈ A ∨ x ∈ B) ∧ φ))
7		andir 838	. . . . . 6 ⊢ (((x ∈ A ∨ x ∈ B) ∧ φ) ↔ ((x ∈ A ∧ φ) ∨ (x ∈ B ∧ φ)))
8	6, 7	bitri 240	. . . . 5 ⊢ ((x ∈ (A ∪ B) ∧ φ) ↔ ((x ∈ A ∧ φ) ∨ (x ∈ B ∧ φ)))
9	8	abbii 2466	. . . 4 ⊢ {x ∣ (x ∈ (A ∪ B) ∧ φ)} = {x ∣ ((x ∈ A ∧ φ) ∨ (x ∈ B ∧ φ))}
10		unab 3522	. . . 4 ⊢ ({x ∣ (x ∈ A ∧ φ)} ∪ {x ∣ (x ∈ B ∧ φ)}) = {x ∣ ((x ∈ A ∧ φ) ∨ (x ∈ B ∧ φ))}
11	9, 10	eqtr4i 2376	. . 3 ⊢ {x ∣ (x ∈ (A ∪ B) ∧ φ)} = ({x ∣ (x ∈ A ∧ φ)} ∪ {x ∣ (x ∈ B ∧ φ)})
12	4, 11	eqtr4i 2376	. 2 ⊢ ({x ∈ A ∣ φ} ∪ {x ∈ B ∣ φ}) = {x ∣ (x ∈ (A ∪ B) ∧ φ)}
13	1, 12	eqtr4i 2376	1 ⊢ {x ∈ (A ∪ B) ∣ φ} = ({x ∈ A ∣ φ} ∪ {x ∈ B ∣ φ})

Syntax hints:   ∨ wo 357   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {cab 2339  {crab 2619   ∪ cun 3208

This theorem was proved from 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 theorem 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-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-un 3215

This theorem is referenced by: (None)