Theorem unrab 4304

Description: Union of two restricted class abstractions. (Contributed by NM, 25-Mar-2004.)

Assertion

Ref	Expression
unrab	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}

Proof of Theorem unrab

Step	Hyp	Ref	Expression
1		df-rab 3433	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		df-rab 3433	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
3	1, 2	uneq12i 4160	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
4		df-rab 3433	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))}
5		unab 4297	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓))}
6		andi 1006	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓)))
7	6	abbii 2802	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓))}
8	5, 7	eqtr4i 2763	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))}
9	4, 8	eqtr4i 2763	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
10	3, 9	eqtr4i 2763	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}

Syntax hints:   ∧ wa 396   ∨ wo 845   = wceq 1541   ∈ wcel 2106  {cab 2709  {crab 3432   ∪ cun 3945

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-12 2171  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-rab 3433  df-v 3476  df-un 3952

This theorem is referenced by:  rabxm  4385  kmlem3  10143  hashbclem  14407  phiprmpw  16705  efgsfo  19601  dsmmacl  21287  rrxmvallem  24912  mumul  26674  ppiub  26696  lgsquadlem2  26873  lrold  27380  edglnl  28392  numclwwlk3lem2lem  29625  zarclsun  32838  hasheuni  33071  measvuni  33200  aean  33230  subfacp1lem6  34164  lineunray  35107  cnambfre  36524  itg2addnclem2  36528  iblabsnclem  36539  orrabdioph  41504  sqrtcvallem1  42367  undisjrab  43050  mndpsuppss  47000