Theorem dfiunv2 5000

Description: Define double indexed union. (Contributed by FL, 6-Nov-2013.)

Assertion

Ref	Expression
dfiunv2	⊢ ∪ 𝑥 ∈ 𝐴 ∪ 𝑦 ∈ 𝐵 𝐶 = {𝑧 ∣ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑧 ∈ 𝐶}

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧   𝑧,𝐴   𝑧,𝐵   𝑧,𝐶

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

Proof of Theorem dfiunv2

Dummy variable 𝑤 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-iun 4960	. . . 4 ⊢ ∪ 𝑦 ∈ 𝐵 𝐶 = {𝑤 ∣ ∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶}
2	1	a1i 11	. . 3 ⊢ (𝑥 ∈ 𝐴 → ∪ 𝑦 ∈ 𝐵 𝐶 = {𝑤 ∣ ∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶})
3	2	iuneq2i 4980	. 2 ⊢ ∪ 𝑥 ∈ 𝐴 ∪ 𝑦 ∈ 𝐵 𝐶 = ∪ 𝑥 ∈ 𝐴 {𝑤 ∣ ∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶}
4		df-iun 4960	. 2 ⊢ ∪ 𝑥 ∈ 𝐴 {𝑤 ∣ ∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶} = {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 ∈ {𝑤 ∣ ∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶}}
5		vex 3465	. . . . 5 ⊢ 𝑧 ∈ V
6		eleq1w 2852	. . . . . 6 ⊢ (𝑤 = 𝑧 → (𝑤 ∈ 𝐶 ↔ 𝑧 ∈ 𝐶))
7	6	rexbidv 3195	. . . . 5 ⊢ (𝑤 = 𝑧 → (∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶 ↔ ∃𝑦 ∈ 𝐵 𝑧 ∈ 𝐶))
8	5, 7	elab 3645	. . . 4 ⊢ (𝑧 ∈ {𝑤 ∣ ∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶} ↔ ∃𝑦 ∈ 𝐵 𝑧 ∈ 𝐶)
9	8	rexbii 3118	. . 3 ⊢ (∃𝑥 ∈ 𝐴 𝑧 ∈ {𝑤 ∣ ∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶} ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑧 ∈ 𝐶)
10	9	abbii 2836	. 2 ⊢ {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 ∈ {𝑤 ∣ ∃𝑦 ∈ 𝐵 𝑤 ∈ 𝐶}} = {𝑧 ∣ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑧 ∈ 𝐶}
11	3, 4, 10	3eqtri 2796	1 ⊢ ∪ 𝑥 ∈ 𝐴 ∪ 𝑦 ∈ 𝐵 𝐶 = {𝑧 ∣ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑧 ∈ 𝐶}

Syntax hints:   = wceq 1567   ∈ wcel 2149  {cab 2747  ∃wrex 3095  ∪ ciun 4958

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rex 3096  df-v 3463  df-ss 3928  df-iun 4960

This theorem is referenced by:  wspniunwspnon  30213  fusgr2wsp2nb  30626