Theorem inrab2 4310

Description: Intersection with a restricted class abstraction. (Contributed by NM, 19-Nov-2007.)

Assertion

Ref	Expression
inrab2	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑}

Distinct variable group:   𝑥,𝐵

Allowed substitution hints:   𝜑(𝑥)   𝐴(𝑥)

Proof of Theorem inrab2

Step	Hyp	Ref	Expression
1		df-rab 3431	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		abid1 2866	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝑥 ∈ 𝐵}
3	1, 2	ineq12i 4212	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵})
4		df-rab 3431	. . 3 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)}
5		inab 4302	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵)}
6		elin 3965	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
7	6	anbi1i 622	. . . . . 6 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑))
8		an32 644	. . . . . 6 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵))
9	7, 8	bitri 274	. . . . 5 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵))
10	9	abbii 2798	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵)}
11	5, 10	eqtr4i 2759	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵}) = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)}
12	4, 11	eqtr4i 2759	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵})
13	3, 12	eqtr4i 2759	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑}

Syntax hints:   ∧ wa 394   = wceq 1533   ∈ wcel 2098  {cab 2705  {crab 3430   ∩ cin 3948

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2706  df-cleq 2720  df-clel 2806  df-rab 3431  df-v 3475  df-in 3956

This theorem is referenced by:  iooval2  13397  fzval2  13527  smuval2  16464  smueqlem  16472  dfphi2  16750  ordtrest  23126  ordtrest2lem  23127  rspectopn  33501  ordtrestNEW  33555  ordtrest2NEWlem  33556  itg2addnclem2  37178  isubgr0uhgr  47235  dmatALTbas  47547