
theorem ND1: for k,n be non zero Nat
  holds k mod n = 0 implies (k - 1) mod n = n - 1
  proof
    let k,n be non zero Nat;
    assume k mod n = 0; then
    n divides k by PEPIN:6; then
    consider m be Nat such that
    A3: k = n*m by NAT_D:def 3;
    A4: (n - 1) + 1 > (n - 1) + 0 by XREAL_1:6;
    reconsider m as non zero Nat by A3;
    reconsider d = m - 1 as Nat;
    (d*n + (n - 1)) mod n = (n - 1) mod n by NAT_D:21
      .= n - 1 by A4;
    hence thesis by A3;
  end;
