reserve k,m,n for Nat,
  i1,i2,i3 for Integer,
  e for set;

theorem
  k <> 0 implies (n+1) mod k = 0 or (n+1) mod k = (n mod k) + 1
proof
  assume
A1: k <> 0;
  then k >= 1 by NAT_1:14;
  then reconsider K = k - 1 as Element of NAT by INT_1:3,XREAL_1:48;
  n mod k < k - 1 + 1 by A1,NAT_D:1;
  then
A2: n mod k <= K by NAT_1:13;
  per cases by A2,XXREAL_0:1;
  suppose
    n mod k < k - 1;
    hence thesis by Th2;
  end;
  suppose
    n mod k = k - 1;
    hence thesis by Th1;
  end;
end;
