Theorem inrab2 4336

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 3444	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		abid1 2881	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝑥 ∈ 𝐵}
3	1, 2	ineq12i 4239	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵})
4		df-rab 3444	. . 3 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)}
5		inab 4328	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵)}
6		elin 3992	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
7	6	anbi1i 623	. . . . . 6 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑))
8		an32 645	. . . . . 6 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵))
9	7, 8	bitri 275	. . . . 5 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵))
10	9	abbii 2812	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵)}
11	5, 10	eqtr4i 2771	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵}) = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)}
12	4, 11	eqtr4i 2771	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵})
13	3, 12	eqtr4i 2771	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑}

Syntax hints:   ∧ wa 395   = wceq 1537   ∈ wcel 2108  {cab 2717  {crab 3443   ∩ cin 3975

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-in 3983

This theorem is referenced by:  iooval2  13440  fzval2  13570  smuval2  16528  smueqlem  16536  dfphi2  16821  ordtrest  23231  ordtrest2lem  23232  rspectopn  33813  ordtrestNEW  33867  ordtrest2NEWlem  33868  itg2addnclem2  37632  isubgr0uhgr  47743  dmatALTbas  48130