reserve i, j, k, l, m, n, t for Nat;

theorem Th24:
  n <= m implies n div k <= m div k
proof
  assume that
A1: n <= m and
A2: n div k > m div k;
  set r = (n div k) -' (m div k);
A3: r = (n div k) - (m div k) by A2,XREAL_1:233;
  then r > (m div k) - (m div k) by A2,XREAL_1:9;
  then r >= 0 + 1 by NAT_1:13;
  then k * r >= k * 1 by XREAL_1:64;
  then
A4: k * r + k * (m div k) >= k + k * (m div k) by XREAL_1:6;
  per cases;
  suppose
A5: k <> 0;
    then ex t2 be Nat st m = k * (m div k) + t2 & t2 < k by NAT_D:def 1;
    then m < k + k * (m div k) by XREAL_1:6;
    then
    m - (k * (n div k)) < (k + k * (m div k)) - (k + k * (m div k)) by A3,A4,
XREAL_1:14;
    then
A6: m - (k * (n div k)) < 0;
    ex t1 be Nat st n = k * (n div k) + t1 & t1 < k by A5,NAT_D:def 1;
    then k * (n div k) <= n by NAT_1:11;
    then m - n < (k * (n div k)) - (k * (n div k)) by A6,XREAL_1:13;
    then m < 0 + n by XREAL_1:19;
    hence contradiction by A1;
  end;
  suppose
    k = 0;
    hence contradiction by A2;
  end;
end;
