Theorem rabun2 4203

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 3062	. 2 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)}
2		df-rab 3062	. . . 4 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
3		df-rab 3062	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
4	2, 3	uneq12i 4052	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
5		elun 4040	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
6	5	anbi1i 627	. . . . . 6 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ 𝜑))
7		andir 1008	. . . . . 6 ⊢ (((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑)))
8	6, 7	bitri 278	. . . . 5 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑)))
9	8	abbii 2803	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑))}
10		unab 4189	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑))}
11	9, 10	eqtr4i 2764	. . 3 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
12	4, 11	eqtr4i 2764	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑}) = {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)}
13	1, 12	eqtr4i 2764	1 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑} = ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑})

Syntax hints:   ∧ wa 399   ∨ wo 846   = wceq 1542   ∈ wcel 2113  {cab 2716  {crab 3057   ∪ cun 3842

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1916  ax-6 1974  ax-7 2019  ax-8 2115  ax-9 2123  ax-10 2144  ax-12 2178  ax-ext 2710

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 847  df-tru 1545  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2717  df-cleq 2730  df-clel 2811  df-rab 3062  df-v 3400  df-un 3849

This theorem is referenced by:  fnsuppres  7887  lfinun  22277  vtxdginducedm1  27485