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Theorem rabun2 3210
Description: Abstraction restricted to a union. (Contributed by Stefan O'Rear, 5-Feb-2015.)
Assertion
Ref Expression
rabun2 {x (AB) ∣ φ} = ({x Aφ} ∪ {x Bφ})

Proof of Theorem rabun2
StepHypRef Expression
1 df-rab 2309 . 2 {x (AB) ∣ φ} = {x ∣ (x (AB) φ)}
2 df-rab 2309 . . . 4 {x Aφ} = {x ∣ (x A φ)}
3 df-rab 2309 . . . 4 {x Bφ} = {x ∣ (x B φ)}
42, 3uneq12i 3089 . . 3 ({x Aφ} ∪ {x Bφ}) = ({x ∣ (x A φ)} ∪ {x ∣ (x B φ)})
5 elun 3078 . . . . . . 7 (x (AB) ↔ (x A x B))
65anbi1i 431 . . . . . 6 ((x (AB) φ) ↔ ((x A x B) φ))
7 andir 731 . . . . . 6 (((x A x B) φ) ↔ ((x A φ) (x B φ)))
86, 7bitri 173 . . . . 5 ((x (AB) φ) ↔ ((x A φ) (x B φ)))
98abbii 2150 . . . 4 {x ∣ (x (AB) φ)} = {x ∣ ((x A φ) (x B φ))}
10 unab 3198 . . . 4 ({x ∣ (x A φ)} ∪ {x ∣ (x B φ)}) = {x ∣ ((x A φ) (x B φ))}
119, 10eqtr4i 2060 . . 3 {x ∣ (x (AB) φ)} = ({x ∣ (x A φ)} ∪ {x ∣ (x B φ)})
124, 11eqtr4i 2060 . 2 ({x Aφ} ∪ {x Bφ}) = {x ∣ (x (AB) φ)}
131, 12eqtr4i 2060 1 {x (AB) ∣ φ} = ({x Aφ} ∪ {x Bφ})
Colors of variables: wff set class
Syntax hints:   wa 97   wo 628   = wceq 1242   wcel 1390  {cab 2023  {crab 2304  cun 2909
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-rab 2309  df-v 2553  df-un 2916
This theorem is referenced by: (None)
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