
theorem ND4:
  for n,k be non zero Nat holds n mod k > 0 implies n div k = (n - 1) div k
  proof
    let n,k be non zero Nat;
    assume
    A1: n mod k > 0;
    n = (n div k)*k + (n mod k) &
      (n - 1) = ((n - 1) div k)*k + ((n - 1) mod k) by NAT_D:2; then
    (n div k)*k/k  = (((n - 1) div k)*k + ((n - 1) mod k) + 1 - (n mod k))/k
    .= (((n - 1) div k)*k + ((n mod k) - 1) + 1 - (n mod k))/k by A1,ND3
    .= (n - 1) div k by XCMPLX_1:89;
    hence thesis by XCMPLX_1:89;
  end;
