Theorem rabun2 4253

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 3306	. 2 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)}
2		df-rab 3306	. . . 4 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
3		df-rab 3306	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
4	2, 3	uneq12i 4101	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
5		elun 4089	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) ↔ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵))
6	5	anbi1i 625	. . . . . 6 ⊢ ((𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ∧ 𝜑))
7		andir 1007	. . . . . 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 2806	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑))}
10		unab 4238	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐵 ∧ 𝜑))}
11	9, 10	eqtr4i 2767	. . 3 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
12	4, 11	eqtr4i 2767	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑}) = {𝑥 ∣ (𝑥 ∈ (𝐴 ∪ 𝐵) ∧ 𝜑)}
13	1, 12	eqtr4i 2767	1 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑} = ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜑})

Syntax hints:   ∧ wa 397   ∨ wo 845   = wceq 1539   ∈ wcel 2104  {cab 2713  {crab 3303   ∪ cun 3890

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 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-10 2135  ax-12 2169  ax-ext 2707

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 846  df-tru 1542  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2714  df-cleq 2728  df-clel 2814  df-rab 3306  df-v 3439  df-un 3897

This theorem is referenced by:  fnsuppres  8038  lfinun  22725  vtxdginducedm1  27959