theorem Th1:
  m + k <= i & i <= n + k implies ex mn st mn + k = i & m <= mn & mn <= n
proof
  assume that
A1: m + k <= i and
A2: i <= n + k;
  m + k <= m + i by A1,XREAL_1:38;
  then k <= i by XREAL_1:6;
  then reconsider mn = i - k as Nat by NAT_1:21;
  take mn;
  thus mn + k = i;
  m + k <= mn + k by A1;
  hence thesis by A2,XREAL_1:6;
end;
