
theorem MDF:
  for n,k be Nat holds n! divides (n+k)!
  proof
    let n,k be Nat;
    defpred P[Nat] means n! divides (n+$1)!;
    A1: P[0];
    A2: for m be Nat st P[m] holds P[m + 1]
    proof
      let m be Nat such that
      B1: n! divides (n+m)!;
      (n+m+1)! = (n + m)!*(n + m + 1) by NEWTON:15; then
      (n + m)! divides (n + m + 1)!;
      hence thesis by B1,INT_2:9;
    end;
    for c be Nat holds P[c] from NAT_1:sch 2(A1,A2);
    hence thesis;
  end;
