Theorem inrab2 4066

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 3064	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		abid1 2887	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝑥 ∈ 𝐵}
3	1, 2	ineq12i 3976	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵})
4		df-rab 3064	. . 3 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)}
5		inab 4061	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵)}
6		elin 3960	. . . . . . 7 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
7	6	anbi1i 617	. . . . . 6 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑))
8		an32 636	. . . . . 6 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵))
9	7, 8	bitri 266	. . . . 5 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵))
10	9	abbii 2882	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝑥 ∈ 𝐵)}
11	5, 10	eqtr4i 2790	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵}) = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ 𝜑)}
12	4, 11	eqtr4i 2790	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ 𝑥 ∈ 𝐵})
13	3, 12	eqtr4i 2790	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ 𝜑}

Syntax hints:   ∧ wa 384   = wceq 1652   ∈ wcel 2155  {cab 2751  {crab 3059   ∩ cin 3733

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1890  ax-4 1904  ax-5 2005  ax-6 2070  ax-7 2105  ax-9 2164  ax-10 2183  ax-11 2198  ax-12 2211  ax-13 2352  ax-ext 2743

This theorem depends on definitions:  df-bi 198  df-an 385  df-or 874  df-tru 1656  df-ex 1875  df-nf 1879  df-sb 2063  df-clab 2752  df-cleq 2758  df-clel 2761  df-nfc 2896  df-rab 3064  df-v 3352  df-in 3741

This theorem is referenced by:  iooval2  12415  fzval2  12541  smuval2  15499  smueqlem  15507  dfphi2  15772  ordtrest  21300  ordtrest2lem  21301  ordtrestNEW  30435  ordtrest2NEWlem  30436  itg2addnclem2  33906  dmatALTbas  42883