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Theorem rabun2 3414
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 2464 . 2 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
2 df-rab 2464 . . . 4 {𝑥𝐴𝜑} = {𝑥 ∣ (𝑥𝐴𝜑)}
3 df-rab 2464 . . . 4 {𝑥𝐵𝜑} = {𝑥 ∣ (𝑥𝐵𝜑)}
42, 3uneq12i 3287 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
5 elun 3276 . . . . . . 7 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
65anbi1i 458 . . . . . 6 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝑥𝐵) ∧ 𝜑))
7 andir 819 . . . . . 6 (((𝑥𝐴𝑥𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
86, 7bitri 184 . . . . 5 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
98abbii 2293 . . . 4 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
10 unab 3402 . . . 4 ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)}) = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
119, 10eqtr4i 2201 . . 3 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
124, 11eqtr4i 2201 . 2 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
131, 12eqtr4i 2201 1 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑})
Colors of variables: wff set class
Syntax hints:  wa 104  wo 708   = wceq 1353  wcel 2148  {cab 2163  {crab 2459  cun 3127
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-rab 2464  df-v 2739  df-un 3133
This theorem is referenced by:  ssfirab  6932
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