Theorem unrab 4306

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 3434	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		df-rab 3434	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
3	1, 2	uneq12i 4162	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
4		df-rab 3434	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))}
5		unab 4299	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓))}
6		andi 1007	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓)))
7	6	abbii 2803	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓))}
8	5, 7	eqtr4i 2764	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))}
9	4, 8	eqtr4i 2764	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
10	3, 9	eqtr4i 2764	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}

Syntax hints:   ∧ wa 397   ∨ wo 846   = wceq 1542   ∈ wcel 2107  {cab 2710  {crab 3433   ∪ cun 3947

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-10 2138  ax-12 2172  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-rab 3434  df-v 3477  df-un 3954

This theorem is referenced by:  rabxm  4387  kmlem3  10147  hashbclem  14411  phiprmpw  16709  efgsfo  19607  dsmmacl  21296  rrxmvallem  24921  mumul  26685  ppiub  26707  lgsquadlem2  26884  lrold  27391  edglnl  28403  numclwwlk3lem2lem  29636  zarclsun  32850  hasheuni  33083  measvuni  33212  aean  33242  subfacp1lem6  34176  lineunray  35119  cnambfre  36536  itg2addnclem2  36540  iblabsnclem  36551  orrabdioph  41519  sqrtcvallem1  42382  undisjrab  43065  mndpsuppss  47047