Theorem inrab2 4252

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 3393	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		abid1 2876	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝑥 ∈ 𝐵}
3	1, 2	ineq12i 4154	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵})
4		df-rab 3393	. . 3 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)}
5		inab 4244	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵)}
6		elin 3906	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
7	6	anbi1i 630	. . . . . 6 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑))
8		an32 652	. . . . . 6 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵))
9	7, 8	bitri 276	. . . . 5 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵))
10	9	abbii 2807	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵)}
11	5, 10	eqtr4i 2766	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵}) = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)}
12	4, 11	eqtr4i 2766	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵})
13	3, 12	eqtr4i 2766	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑}

Syntax hints:   ∧ wa 396   = wceq 1547   ∈ wcel 2119  {cab 2718  {crab 3392   ∩ cin 3889

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-rab 3393  df-v 3434  df-in 3897

This theorem is referenced by:  iooval2  13329  fzval2  13462  smuval2  16449  smueqlem  16457  dfphi2  16742  ordtrest  23192  ordtrest2lem  23193  rspectopn  34058  ordtrestNEW  34112  ordtrest2NEWlem  34113  itg2addnclem2  38046  isubgr0uhgr  48371  dmatALTbas  48899