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Theorem rabun2 3861
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 2901 . 2 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
2 df-rab 2901 . . . 4 {𝑥𝐴𝜑} = {𝑥 ∣ (𝑥𝐴𝜑)}
3 df-rab 2901 . . . 4 {𝑥𝐵𝜑} = {𝑥 ∣ (𝑥𝐵𝜑)}
42, 3uneq12i 3723 . . 3 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
5 elun 3711 . . . . . . 7 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
65anbi1i 726 . . . . . 6 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝑥𝐵) ∧ 𝜑))
7 andir 907 . . . . . 6 (((𝑥𝐴𝑥𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
86, 7bitri 262 . . . . 5 ((𝑥 ∈ (𝐴𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑)))
98abbii 2722 . . . 4 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
10 unab 3849 . . . 4 ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)}) = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵𝜑))}
119, 10eqtr4i 2631 . . 3 {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)} = ({𝑥 ∣ (𝑥𝐴𝜑)} ∪ {𝑥 ∣ (𝑥𝐵𝜑)})
124, 11eqtr4i 2631 . 2 ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑}) = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ 𝜑)}
131, 12eqtr4i 2631 1 {𝑥 ∈ (𝐴𝐵) ∣ 𝜑} = ({𝑥𝐴𝜑} ∪ {𝑥𝐵𝜑})
Colors of variables: wff setvar class
Syntax hints:  wo 381  wa 382   = wceq 1474  wcel 1976  {cab 2592  {crab 2896  cun 3534
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-11 2020  ax-12 2032  ax-13 2229  ax-ext 2586
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867  df-clab 2593  df-cleq 2599  df-clel 2602  df-nfc 2736  df-rab 2901  df-v 3171  df-un 3541
This theorem is referenced by:  fnsuppres  7183  lfinun  21077
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