
theorem XLMOD01:
  for k,m be Nat st m <> 0 & (k+1) mod m <> 0 holds (k+1) div m = k div m
proof
  let k,m be Nat;
  assume
C1: m <> 0 & (k+1) mod m <> 0;
  k+1 = ((k+1) div m )*m+((k+1) mod m) by INT_1:59,C1
    .= ((k+1) div m )*m+((k mod m)+1) by XLMOD02,C1;
  then
P1: ((k+1) div m)*m+(k mod m)-(k mod m) = (k div m )*m+(k mod m)-(k mod m)
  by INT_1:59,C1;
  thus ((k+1) div m) = (k div m)*m / m by XCMPLX_1:89,C1,P1
    .= (k div m) by XCMPLX_1:89,C1;
end;
