
theorem Th15:
  for n being Nat st n > 1 holds Catalan (n) < Catalan (n + 1)
proof
  let n be Nat;
  set a = (2 * n -' 2)!;
  set b = (2 * n)!;
A1: Catalan (n + 1) = (2*n)! / (n! * ((n+1)!)) by Th14;
  assume
A2: n > 1;
  then n -' 1 + 1 = n by XREAL_1:235;
  then
A3: (a * n * (n + 1)) / (n! * ((n + 1)!)) = (a * n * (n + 1)) / (n * ((n -'
  1)!)* ((n + 1)!)) by NEWTON:15
    .= (n * (a * (n + 1))) / (n * (((n -' 1)!)* ((n + 1)!)))
    .= (a * (n + 1)) /(((n -' 1)!) * ((n + 1)!)) by A2,XCMPLX_1:91
    .= (a * (n + 1)) / (((n -' 1)!) * ((n + 1) * (n!))) by NEWTON:15
    .= (a * (n + 1)) / (((n -' 1)!) * (n!) * (n + 1))
    .= a / ((n -' 1)! * (n!)) by XCMPLX_1:91;
  n! > 0 & (n + 1)! > 0 by NEWTON:17;
  then n! * ((n + 1)!) > 0 * (n!) by XREAL_1:68;
  then (a * n * (n + 1)) / (n! * ((n + 1)!)) < b / (n! * ((n + 1)!)) by A2,Th6,
XREAL_1:74;
  hence thesis by A2,A1,A3,Th8;
end;
