
theorem Th14:
  for k being Nat holds Catalan (k + 1) = (2*k)! / (k! * ((k+1)!))
proof
  let k be Nat;
  reconsider l = 2*k - k as Nat;
  l = k & 1*k <= 2*k by XREAL_1:64;
  then
A1: (2*k) choose k = (2*k!)/((k!) * (k!)) by NEWTON:def 3;
  2*k + 2 -' 2 = 2*k & k + 1 -' 1 = k by NAT_D:34;
  then Catalan (k+1) = (2*k!) / ((k!) * (k!) * (k+1)) by A1,XCMPLX_1:78
    .= (2*k!) / ((k!) * ((k!) * (k+1)))
    .= (2*k!) / ((k!) * ((k+1)!)) by NEWTON:15;
  hence thesis;
end;
