Theorem rabun2 4290

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 3409	. 2 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)}
2		df-rab 3409	. . . 4 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
3		df-rab 3409	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
4	2, 3	uneq12i 4132	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
5		elun 4119	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
6	5	anbi1i 624	. . . . . 6 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ 𝜑))
7		andir 1010	. . . . . 6 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑)))
8	6, 7	bitri 275	. . . . 5 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑)))
9	8	abbii 2797	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑))}
10		unab 4274	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑))}
11	9, 10	eqtr4i 2756	. . 3 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
12	4, 11	eqtr4i 2756	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑}) = {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)}
13	1, 12	eqtr4i 2756	1 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑} = ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑})

Syntax hints:   ∧ wa 395   ∨ wo 847   = wceq 1540   ∈ wcel 2109  {cab 2708  {crab 3408   ∪ cun 3915

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-12 2178  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-rab 3409  df-v 3452  df-un 3922

This theorem is referenced by:  fnsuppres  8173  lfinun  23419  vtxdginducedm1  29478