
theorem XLMOD01X:
  for k,m be Nat st m <> 0 & (k+1) mod m = 0
  holds (k+1) div m = (k div m)+1
proof
  let k,m be Nat;
  assume
C1: m <> 0 & (k+1) mod m = 0;
  then
P3: k mod m = m-1 by XLMOD02X;
P4: k+1 = ((k+1) div m)*m+((k+1) mod m) by INT_1:59,C1
    .= ((k+1) div m)*m by C1;
P5: k = (k div m )*m+(k mod m) by INT_1:59,C1
    .= (k div m)*m+m-1 by P3;
  thus ((k+1) div m) = ((k div m)+1)*m / m by XCMPLX_1:89,C1,P4,P5
    .= ((k div m)+1) by XCMPLX_1:89,C1;
end;
