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Theorem rabun2 4110
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 3102 . 2 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
2 df-rab 3102 . . . 4 {𝑥𝐴𝜑} = {𝑥 ∣ (𝑥𝐴𝜑)}
3 df-rab 3102 . . . 4 {𝑥𝐵𝜑} = {𝑥 ∣ (𝑥𝐵𝜑)}
42, 3uneq12i 3967 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
5 elun 3955 . . . . . . 7 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
65anbi1i 618 . . . . . 6 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝑥𝐵) ∧ 𝜑))
7 andir 1032 . . . . . 6 (((𝑥𝐴𝑥𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
86, 7bitri 267 . . . . 5 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
98abbii 2920 . . . 4 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
10 unab 4098 . . . 4 ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)}) = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
119, 10eqtr4i 2828 . . 3 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
124, 11eqtr4i 2828 . 2 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
131, 12eqtr4i 2828 1 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑})
Colors of variables: wff setvar class
Syntax hints:  wa 385  wo 874   = wceq 1653  wcel 2157  {cab 2789  {crab 3097  cun 3771
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2379  ax-ext 2781
This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2790  df-cleq 2796  df-clel 2799  df-nfc 2934  df-rab 3102  df-v 3391  df-un 3778
This theorem is referenced by:  fnsuppres  7564  lfinun  21661  vtxdginducedm1  26797
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