
theorem Th18:
  for i,j,k being Integer holds i gcd (j gcd k) = (i gcd j) gcd k
proof
  let i,j,k be Integer;
  per cases;
  suppose
    i = 0 & j = 0 & k = 0;
    hence thesis;
  end;
  suppose
A1: i <> 0 or j <> 0 or k <> 0;
A2: now
      assume i gcd (j gcd k) = -((i gcd j) gcd k);
      then (-((i gcd j) gcd k)) * (-1) <= 0 * (-1);
      then
A3:   (i gcd j) gcd k = 0;
      then i gcd j = 0 by INT_2:5;
      hence contradiction by A1,A3,INT_2:5;
    end;
A4: i gcd (j gcd k) divides i by INT_2:21;
A5: (i gcd j) gcd k divides k by INT_2:21;
A6: (i gcd j) gcd k divides i gcd j by INT_2:21;
    i gcd j divides j by INT_2:21;
    then (i gcd j) gcd k divides j by A6,INT_2:9;
    then
A7: (i gcd j) gcd k divides j gcd k by A5,INT_2:22;
    i gcd j divides i by INT_2:21;
    then (i gcd j) gcd k divides i by A6,INT_2:9;
    then
A8: (i gcd j) gcd k divides i gcd (j gcd k) by A7,INT_2:22;
A9: i gcd (j gcd k) divides j gcd k by INT_2:21;
    j gcd k divides j by INT_2:21;
    then i gcd (j gcd k) divides j by A9,INT_2:9;
    then
A10: i gcd (j gcd k) divides i gcd j by A4,INT_2:22;
    j gcd k divides k by INT_2:21;
    then i gcd (j gcd k) divides k by A9,INT_2:9;
    then i gcd (j gcd k) divides (i gcd j) gcd k by A10,INT_2:22;
    hence thesis by A8,A2,INT_2:11;
  end;
end;
