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Theorem rabun2 3360
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 2426 . 2 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
2 df-rab 2426 . . . 4 {𝑥𝐴𝜑} = {𝑥 ∣ (𝑥𝐴𝜑)}
3 df-rab 2426 . . . 4 {𝑥𝐵𝜑} = {𝑥 ∣ (𝑥𝐵𝜑)}
42, 3uneq12i 3233 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
5 elun 3222 . . . . . . 7 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
65anbi1i 454 . . . . . 6 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝑥𝐵) ∧ 𝜑))
7 andir 809 . . . . . 6 (((𝑥𝐴𝑥𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
86, 7bitri 183 . . . . 5 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
98abbii 2256 . . . 4 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
10 unab 3348 . . . 4 ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)}) = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
119, 10eqtr4i 2164 . . 3 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
124, 11eqtr4i 2164 . 2 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
131, 12eqtr4i 2164 1 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑})
Colors of variables: wff set class
Syntax hints:  wa 103  wo 698   = wceq 1332  wcel 1481  {cab 2126  {crab 2421  cun 3074
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122
This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-rab 2426  df-v 2691  df-un 3080
This theorem is referenced by:  ssfirab  6830
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