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Theorem iunun 5058
Description: Separate a union in an indexed union. (Contributed by NM, 27-Dec-2004.) (Proof shortened by Mario Carneiro, 17-Nov-2016.)
Assertion
Ref Expression
iunun ∪ 𝑥 ∈ 𝐴 (𝐵 ∪ 𝐶) = (∪ 𝑥 ∈ 𝐴 𝐵 ∪ ∪ 𝑥 ∈ 𝐴 𝐶)

Proof of Theorem iunun
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 r19.43 3134 . . . 4 (∃𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∨ 𝑦 ∈ 𝐶) ↔ (∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 ∨ ∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐶))
2 elun 4155 . . . . 5 (𝑦 ∈ (𝐵 ∪ 𝐶) ↔ (𝑦 ∈ 𝐵 ∨ 𝑦 ∈ 𝐶))
32rexbii 3113 . . . 4 (∃𝑥 ∈ 𝐴 𝑦 ∈ (𝐵 ∪ 𝐶) ↔ ∃𝑥 ∈ 𝐴 (𝑦 ∈ 𝐵 ∨ 𝑦 ∈ 𝐶))
4 eliun 4960 . . . . 5 (𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐵 ↔ ∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐵)
5 eliun 4960 . . . . 5 (𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐶 ↔ ∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐶)
64, 5orbi12i 928 . . . 4 ((𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐵 ∨ 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐶) ↔ (∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 ∨ ∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐶))
71, 3, 63bitr4i 306 . . 3 (∃𝑥 ∈ 𝐴 𝑦 ∈ (𝐵 ∪ 𝐶) ↔ (𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐵 ∨ 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐶))
8 eliun 4960 . . 3 (𝑦 ∈ ∪ 𝑥 ∈ 𝐴 (𝐵 ∪ 𝐶) ↔ ∃𝑥 ∈ 𝐴 𝑦 ∈ (𝐵 ∪ 𝐶))
9 elun 4155 . . 3 (𝑦 ∈ (∪ 𝑥 ∈ 𝐴 𝐵 ∪ ∪ 𝑥 ∈ 𝐴 𝐶) ↔ (𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐵 ∨ 𝑦 ∈ ∪ 𝑥 ∈ 𝐴 𝐶))
107, 8, 93bitr4i 306 . 2 (𝑦 ∈ ∪ 𝑥 ∈ 𝐴 (𝐵 ∪ 𝐶) ↔ 𝑦 ∈ (∪ 𝑥 ∈ 𝐴 𝐵 ∪ ∪ 𝑥 ∈ 𝐴 𝐶))
1110eqriv 2760 1 ∪ 𝑥 ∈ 𝐴 (𝐵 ∪ 𝐶) = (∪ 𝑥 ∈ 𝐴 𝐵 ∪ ∪ 𝑥 ∈ 𝐴 𝐶)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   ∨ wo 861   = wceq 1570   ∈ wcel 2147  ∃wrex 3090   ∪ cun 4152  ∪ ciun 4956
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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-v 3456  df-un 4154  df-iun 4958
This theorem is used by:  iununi  5064  oarec  8608  comppfsc  23906  uniiccdif  25954  bnj1415  35779  ttciun  37517
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