
theorem Th19:
  for n being Nat st n > 0 holds Catalan (n+1) = 2 * (2
  - (3 / (n + 1))) * Catalan (n)
proof
  let n be Nat;
  assume
A1: n > 0;
  then
A2: n >= 1 + 0 by NAT_1:13;
  then
A3: 2 * (n -' 1) = 2 * (n - 1) by XREAL_1:233
    .= 2 * n - 2 * 1
    .= 2 * n -' 2 by A2,XREAL_1:64,233;
A4: Catalan n = Catalan (n -' 1 + 1) by A2,XREAL_1:235
    .= (2 * (n -' 1))! / ((n -' 1)! * ((n -' 1 + 1)!)) by Th14
    .= (2 * n -' 2)! / ((n -' 1)! * (n!)) by A2,A3,XREAL_1:235;
A5: n -' 1 + 1 = n by A2,XREAL_1:235;
A6: 1 * 2 <= 2 * n by A2,XREAL_1:64;
  then
A7: 2 * n -' 1 = 2 * n - 1 by XREAL_1:233,XXREAL_0:2;
A8: 2 * (2 - (3 / (n + 1))) = 2 * ((2 * (n + 1) / (n + 1) - (3 / (n + 1))))
  by XCMPLX_1:89
    .= 2 * ((2 * (n + 1) - 3) / (n + 1)) by XCMPLX_1:120
    .= (2 * (2 * n - 1)) / (n + 1) by XCMPLX_1:74
    .= ((2 * n -' 1) * 2 * n) / (n * (n + 1)) by A1,A7,XCMPLX_1:91
    .= ((2 * n -' 1) * (2 * n)) / (n * (n + 1));
A9: 2 * n -' 1 + 1 = 2 * n by A6,XREAL_1:235,XXREAL_0:2;
  1 < 2 * n by A6,XXREAL_0:2;
  then
A10: 2 * n -' 2 + 1 = 2 * n -' 1 by Th4;
  Catalan (n + 1) = (2*n)! / (n! * ((n+1)!)) by Th14
    .= ((2*n -' 1)! * (2*n)) / (n! * ((n+1)!)) by A9,NEWTON:15
    .= ((2*n -' 2)! * (2 * n -' 1) * (2*n)) / (n! * ((n+1)!)) by A10,NEWTON:15
    .= ((2*n -' 2)! * (2 * n -' 1) * (2*n)) / (n! * (n! * (n+1))) by NEWTON:15
    .= ((2*n -' 2)! * (2 * n -' 1) * (2*n)) / (n! * ((n -' 1)! * n * (n+1)))
  by A5,NEWTON:15
    .= ((2*n -' 2)! * ((2 * n -' 1) * (2*n))) / (n! * ((n -' 1)!) * (n * (n+
  1)))
    .= (Catalan n) * (((2 * n -' 1) * (2*n)) / (n * (n+1))) by A4,XCMPLX_1:76;
  hence thesis by A8;
end;
