Theorem unrab 3475

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 2517	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		df-rab 2517	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
3	1, 2	uneq12i 3356	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
4		df-rab 2517	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))}
5		unab 3471	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓))}
6		andi 823	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓)))
7	6	abbii 2345	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∨ (𝑥 ∈ 𝐴 ∧ 𝜓))}
8	5, 7	eqtr4i 2253	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∨ 𝜓))}
9	4, 8	eqtr4i 2253	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
10	3, 9	eqtr4i 2253	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}

Syntax hints:   ∧ wa 104   ∨ wo 713   = wceq 1395   ∈ wcel 2200  {cab 2215  {crab 2512   ∪ cun 3195

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-rab 2517  df-v 2801  df-un 3201

This theorem is referenced by:  rabxmdc  3523  phiprmpw  12739  unennn  12963  znnen  12964  lgsquadlem2  15751