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Theorem rabun2 3241
Description: Abstraction restricted to a union. (Contributed by Stefan O'Rear, 5-Feb-2015.)
Assertion
Ref Expression
rabun2 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑})

Proof of Theorem rabun2
StepHypRef Expression
1 df-rab 2330 . 2 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
2 df-rab 2330 . . . 4 {𝑥𝐴𝜑} = {𝑥 ∣ (𝑥𝐴𝜑)}
3 df-rab 2330 . . . 4 {𝑥𝐵𝜑} = {𝑥 ∣ (𝑥𝐵𝜑)}
42, 3uneq12i 3120 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
5 elun 3109 . . . . . . 7 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
65anbi1i 439 . . . . . 6 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝑥𝐵) ∧ 𝜑))
7 andir 741 . . . . . 6 (((𝑥𝐴𝑥𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
86, 7bitri 177 . . . . 5 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
98abbii 2167 . . . 4 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
10 unab 3229 . . . 4 ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)}) = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
119, 10eqtr4i 2077 . . 3 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
124, 11eqtr4i 2077 . 2 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
131, 12eqtr4i 2077 1 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑})
Colors of variables: wff set class
Syntax hints:  wa 101  wo 637   = wceq 1257  wcel 1407  {cab 2040  {crab 2325  cun 2940
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 638  ax-5 1350  ax-7 1351  ax-gen 1352  ax-ie1 1396  ax-ie2 1397  ax-8 1409  ax-10 1410  ax-11 1411  ax-i12 1412  ax-bndl 1413  ax-4 1414  ax-17 1433  ax-i9 1437  ax-ial 1441  ax-i5r 1442  ax-ext 2036
This theorem depends on definitions:  df-bi 114  df-tru 1260  df-nf 1364  df-sb 1660  df-clab 2041  df-cleq 2047  df-clel 2050  df-nfc 2181  df-rab 2330  df-v 2574  df-un 2947
This theorem is referenced by: (None)
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