
theorem Th13:
  for X being set, V being Subset of X holds chi(V,X)"{1} = V &
  chi(V,X)"{0} = X\V
proof
  let X be set;
  let V be Subset of X;
  thus chi(V,X)"{1} = V
  proof
    thus chi(V,X)"{1} c= V
    proof
      let x be object;
       reconsider xx=x as set by TARSKI:1;
      assume
A1:   x in chi(V,X)"{1};
      then chi(V,X).x in {1} by FUNCT_1:def 7;
      then
    chi(V,X).xx = {0} by TARSKI:def 1,CARD_1:49;
       then chi(V,X).xx <> {};
      hence thesis by A1,FUNCT_3:def 3;
    end;
    let x be object;
    assume
A2: x in V;
    then chi(V,X).x = 1 by FUNCT_3:def 3;
    then
A3: chi(V,X).x in {1} by TARSKI:def 1;
    x in X by A2;
    then x in dom chi(V,X) by FUNCT_3:def 3;
    hence thesis by A3,FUNCT_1:def 7;
  end;
  thus chi(V,X)"{0} = X\V
  proof
    thus chi(V,X)"{0} c= X\V
    proof
      let x be object;
A4:   x in V implies chi(V,X).x = 1 by FUNCT_3:def 3;
      assume
A5:   x in chi(V,X)"{0};
      then chi(V,X).x in {0} by FUNCT_1:def 7;
      hence thesis by A4,A5,TARSKI:def 1,XBOOLE_0:def 5;
    end;
    let x be object;
    assume
A6: x in X\V;
    then not x in V by XBOOLE_0:def 5;
    then chi(V,X).x = 0 by A6,FUNCT_3:def 3;
    then
A7: chi(V,X).x in {0} by TARSKI:def 1;
    x in X by A6;
    then x in dom chi(V,X) by FUNCT_3:def 3;
    hence thesis by A7,FUNCT_1:def 7;
  end;
end;
