reserve n, k, r, m, i, j for Nat;

theorem Th14:
  for k being Element of NAT st k <> 0 holds k + m <= n implies m < n
proof
  let k be Element of NAT;
  assume
A1: k <> 0;
  assume
A2: k + m <= n;
  per cases by A2,XXREAL_0:1;
  suppose
    k + m < n;
    hence thesis by NAT_1:12;
  end;
  suppose
A3: k + m = n;
    assume not m < n;
    then m + k >= n + k by XREAL_1:7;
    then n - n >= n + k - n by A3,XREAL_1:9;
    hence contradiction by A1;
  end;
end;
