Theorem rabiun 38489

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 4955	. . . . . 6 ⊢ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ↔ ∃𝑦 ∈ 𝐴 𝑥 ∈ 𝐵)
2	1	anbi1i 636	. . . . 5 ⊢ ((𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑) ↔ (∃𝑦 ∈ 𝐴 𝑥 ∈ 𝐵 ∧ 𝜑))
3		r19.41v 3193	. . . . 5 ⊢ (∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (∃𝑦 ∈ 𝐴 𝑥 ∈ 𝐵 ∧ 𝜑))
4	2, 3	bitr4i 281	. . . 4 ⊢ ((𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑) ↔ ∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑))
5	4	abbii 2828	. . 3 ⊢ {𝑥 ∣ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑)} = {𝑥 ∣ ∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑)}
6		df-rab 3414	. . 3 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∧ 𝜑)}
7		iunab 5010	. . 3 ⊢ ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} = {𝑥 ∣ ∃𝑦 ∈ 𝐴 (𝑥 ∈ 𝐵 ∧ 𝜑)}
8	5, 6, 7	3eqtr4i 2794	. 2 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
9		df-rab 3414	. . . 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 4973	. 2 ⊢ ∪ 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
12	8, 11	eqtr4i 2787	1 ⊢ {𝑥 ∈ ∪ 𝑦 ∈ 𝐴 𝐵 ∣ 𝜑} = ∪ 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝜑}

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {cab 2739  ∃wrex 3087  {crab 3413  ∪ ciun 4951

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-ss 3916  df-iun 4953

This theorem is used by:  itg2addnclem2  38558