Theorem unrab 4270

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 3418	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		df-rab 3418	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
3	1, 2	uneq12i 4122	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
4		df-rab 3418	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))}
5		unab 4263	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓))}
6		andi 1023	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓)))
7	6	abbii 2832	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓))}
8	5, 7	eqtr4i 2791	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))}
9	4, 8	eqtr4i 2791	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
10	3, 9	eqtr4i 2791	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}

Syntax hints:   ∧ wa 400   ∨ wo 860   = wceq 1563   ∈ wcel 2145  {cab 2743  {crab 3417   ∪ cun 3905

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-10 2178  ax-12 2215  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1566  df-ex 1803  df-nf 1807  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3418  df-v 3459  df-un 3912

This theorem is referenced by:  rabxm  4347  kmlem3  10124  hashbclem  14479  phiprmpw  16825  mndpsuppss  18813  efgsfo  19800  dsmmacl  21851  rrxmvallem  25524  mumul  27303  ppiub  27326  lgsquadlem2  27503  lrold  28048  edglnl  29402  numclwwlk3lem2lem  30643  3unrab  32759  zarclsun  34177  hasheuni  34392  measvuni  34521  aean  34551  subfacp1lem6  35548  lineunray  36510  cnambfre  38179  itg2addnclem2  38183  iblabsnclem  38194  orrabdioph  43374  sqrtcvallem1  44219  undisjrab  44880  dfsclnbgr6  48478