Theorem rabiun 36461

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 5002	. . . . . 6 ⊢ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ↔ ∃𝑦 ∈ 𝐴 𝑥 ∈ 𝐵)
2	1	anbi1i 625	. . . . 5 ⊢ ((𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑) ↔ (∃𝑦 ∈ 𝐴 𝑥 ∈ 𝐵 ∧ 𝜑))
3		r19.41v 3189	. . . . 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 2803	. . 3 ⊢ {𝑥 ∣ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑)} = {𝑥 ∣ ∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑)}
6		df-rab 3434	. . 3 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑)}
7		iunab 5055	. . 3 ⊢ ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} = {𝑥 ∣ ∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑)}
8	5, 6, 7	3eqtr4i 2771	. 2 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
9		df-rab 3434	. . . 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 5019	. 2 ⊢ ∪ 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
12	8, 11	eqtr4i 2764	1 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝜑}

Syntax hints:   ∧ wa 397   = wceq 1542   ∈ wcel 2107  {cab 2710  ∃wrex 3071  {crab 3433  ∪ ciun 4998

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2155  ax-12 2172  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-nfc 2886  df-ral 3063  df-rex 3072  df-rab 3434  df-v 3477  df-in 3956  df-ss 3966  df-iun 5000

This theorem is referenced by:  itg2addnclem2  36540