
theorem ND2:
  for k be Nat, n be non zero Nat holds
    (k + 1) mod n = 0 implies (k + 1) div n = (k div n) + 1
proof
  let k be Nat, n be non zero Nat;
  assume
  A1: (k + 1) mod n = 0;
  n >= 1 by NAT_1:14; then
  per cases by XXREAL_0:1;
  suppose n = 1; then
    (k + 1) div n = k + 1 & k div n = k & 1 div n = 1 by NAT_2:4;
    hence thesis;
  end;
  suppose
    B1: n > 1;
    n divides k + 1 by A1,PEPIN:6; then
    reconsider k as non zero Nat by B1,NAT_D:7;
    B2: ((k + 1) - 1) mod n = n - 1 by A1,ND1;
    B3: n mod n = 0;
    B4: ((k div n)*n) div n = (k div n) & ((1*n) div n) = 1 by NAT_D:18;
    (1 + k) div n = (1 + ((k div n)*n + (k mod n))) div n by NAT_D:2
    .= (n + (k div n)*n) div n by B2
    .= (k div n) + 1 by B3,B4,NAT_D:19;
    hence thesis;
  end;
end;
