reserve a,b,i,j,k,l,m,n for Nat;

theorem NCK:
  (n+k) choose k = 1 iff n = 0 or k = 0
  proof
    A1: n <> 0 & k <> 0 implies (n+k) choose k <> 1
    proof
      assume n <> 0 & k <> 0; then
      reconsider m = n-1,l=k-1 as Nat;
     (m+k) choose k >= 1 & (n+l) choose l >= 1 by NAT_1:14; then
      B2: ((k+m) choose k) + ((l+n) choose l) >= 1 + 1 by XREAL_1:7;
      ((m+k)+1) choose (l+1) = ((m+k) choose k) + ((n+l) choose l)
        by NEWTON:22;
      hence thesis by B2;
    end;
    n = 0 or k = 0 implies (n+k) choose (k*1) = 1;
    hence thesis by A1;
  end;
