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Theorem inrab2 3204
Description: Intersection with a restricted class abstraction. (Contributed by NM, 19-Nov-2007.)
Assertion
Ref Expression
inrab2 ({x Aφ} ∩ B) = {x (AB) ∣ φ}
Distinct variable group:   x,B
Allowed substitution hints:   φ(x)   A(x)

Proof of Theorem inrab2
StepHypRef Expression
1 df-rab 2309 . . 3 {x Aφ} = {x ∣ (x A φ)}
2 abid2 2155 . . . 4 {xx B} = B
32eqcomi 2041 . . 3 B = {xx B}
41, 3ineq12i 3130 . 2 ({x Aφ} ∩ B) = ({x ∣ (x A φ)} ∩ {xx B})
5 df-rab 2309 . . 3 {x (AB) ∣ φ} = {x ∣ (x (AB) φ)}
6 inab 3199 . . . 4 ({x ∣ (x A φ)} ∩ {xx B}) = {x ∣ ((x A φ) x B)}
7 elin 3120 . . . . . . 7 (x (AB) ↔ (x A x B))
87anbi1i 431 . . . . . 6 ((x (AB) φ) ↔ ((x A x B) φ))
9 an32 496 . . . . . 6 (((x A x B) φ) ↔ ((x A φ) x B))
108, 9bitri 173 . . . . 5 ((x (AB) φ) ↔ ((x A φ) x B))
1110abbii 2150 . . . 4 {x ∣ (x (AB) φ)} = {x ∣ ((x A φ) x B)}
126, 11eqtr4i 2060 . . 3 ({x ∣ (x A φ)} ∩ {xx B}) = {x ∣ (x (AB) φ)}
135, 12eqtr4i 2060 . 2 {x (AB) ∣ φ} = ({x ∣ (x A φ)} ∩ {xx B})
144, 13eqtr4i 2060 1 ({x Aφ} ∩ B) = {x (AB) ∣ φ}
Colors of variables: wff set class
Syntax hints:   wa 97   = wceq 1242   wcel 1390  {cab 2023  {crab 2304  cin 2910
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-rab 2309  df-v 2553  df-in 2918
This theorem is referenced by:  iooval2  8514  fzval2  8607
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