
theorem XLMOD02X:
  for k,m be Nat st m <> 0 & (k+1) mod m = 0 holds m-1 = (k mod m)
proof
  let k,m be Nat;
  assume
C1: m <> 0 & (k+1) mod m = 0;
  then (k mod m)+1 <= m by NAT_D:1,NAT_1:13;
  then
P1: (k mod m)+1-1 <= m-1 by XREAL_1:9;
P2: (k+1) mod m = ((k mod m)+1) mod m by NAT_D:22;
  assume not k mod m = m-1;
  then k mod m < m-1 by XXREAL_0:1,P1;
  then k mod m+1 < m-1+1 by XREAL_1:8;
  hence contradiction by P2,NAT_D:24,C1;
end;
