Theorem rabiun 36972

Description: Abstraction restricted to an indexed union. (Contributed by Brendan Leahy, 26-Oct-2017.)

Assertion

Ref	Expression
rabiun	⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝜑}

Distinct variable groups:   𝜑,𝑦   𝑥,𝐴   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥)   𝐴(𝑦)   𝐵(𝑥,𝑦)

Proof of Theorem rabiun

Step	Hyp	Ref	Expression
1		eliun 4994	. . . . . 6 ⊢ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ↔ ∃𝑦 ∈ 𝐴 𝑥 ∈ 𝐵)
2	1	anbi1i 623	. . . . 5 ⊢ ((𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑) ↔ (∃𝑦 ∈ 𝐴 𝑥 ∈ 𝐵 ∧ 𝜑))
3		r19.41v 3182	. . . . 5 ⊢ (∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (∃𝑦 ∈ 𝐴 𝑥 ∈ 𝐵 ∧ 𝜑))
4	2, 3	bitr4i 278	. . . 4 ⊢ ((𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑) ↔ ∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑))
5	4	abbii 2796	. . 3 ⊢ {𝑥 ∣ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑)} = {𝑥 ∣ ∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑)}
6		df-rab 3427	. . 3 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑)}
7		iunab 5047	. . 3 ⊢ ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} = {𝑥 ∣ ∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑)}
8	5, 6, 7	3eqtr4i 2764	. 2 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
9		df-rab 3427	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
10	9	a1i 11	. . 3 ⊢ (𝑦 ∈ 𝐴 → {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
11	10	iuneq2i 5011	. 2 ⊢ ∪ 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
12	8, 11	eqtr4i 2757	1 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝜑}

Syntax hints:   ∧ wa 395   = wceq 1533   ∈ wcel 2098  {cab 2703  ∃wrex 3064  {crab 3426  ∪ ciun 4990

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-nfc 2879  df-ral 3056  df-rex 3065  df-rab 3427  df-v 3470  df-in 3950  df-ss 3960  df-iun 4992

This theorem is referenced by:  itg2addnclem2  37051