theorem Th6:
  n divides i implies i div n = i / n
  proof
    per cases;
    suppose n = 0;
      hence thesis;
    end;
    suppose
A1:   n <> 0;
      assume
A2:   n divides i;
      i mod n = 0 by A1,A2,INT_1:62;
      hence thesis by PEPIN:63;
    end;
  end;
