
theorem Th6:
  for f being non empty one-to-one Function
  for X being non empty Subset of bool dom f
  holds rng((.:f)|X) = the set of all f.:x where x is Element of X
proof
  let f be non empty one-to-one Function, X be non empty Subset of bool dom f;
  set S = the set of all f.:x where x is Element of X;
  now
    let y be object;
    hereby
      assume y in S;
      then consider x9 being Element of X such that
        A1: y = f.:x9;
      reconsider x = x9 as object;
      take x;
      x in bool dom f;
      then A2: x in dom(.:f) by FUNCT_3:def 1;
      hence x in dom((.:f)|X) by RELAT_1:57;
      thus y = (.:f).x by A1, A2, FUNCT_3:7
        .= ((.:f)|X).x by FUNCT_1:49;
    end;
    given x being object such that
      A3: x in dom((.:f)|X) & y = ((.:f)|X).x;
    A4: x in dom(.:f) & x in X by A3, RELAT_1:57;
    then reconsider x9 = x as Element of X;
    y = (.:f).x by A3, FUNCT_1:47
      .= f.:x9 by A4, FUNCT_3:7;
    hence y in S;
  end;
  hence thesis by FUNCT_1:def 3;
end;
