
theorem Th9:
  for n being Nat st n > 1 holds Catalan n = 4 * ((2*n
  -' 3) choose (n -' 1)) - ((2*n -' 1) choose (n -' 1))
proof
  let n be Nat;
  assume
A1: n > 1;
  then
A2: n -' 1 <= 2 * n -' 3 by Th1;
A3: 2 * 1 <= 2 * n by A1,XREAL_1:64;
  then
A4: 2 * n -' 2 = 2 * n - 2 by XREAL_1:233;
A5: 1 + 1 <= n by A1,NAT_1:13;
  then
A6: 2 * 2 <= 2 * n by XREAL_1:64;
  then 2 * n -' 3 = 2 * n - 3 by XREAL_1:233,XXREAL_0:2;
  then
A7: 2 * n -' 3 + 1 = 2 * n - 2 .= 2 * n -' 2 by A3,XREAL_1:233;
  (2*n -' 3) - (n -' 1) = (2*n -' 3) - (n - 1) by A1,XREAL_1:233
    .= (2*n - 3) - (n - 1) by A6,XREAL_1:233,XXREAL_0:2
    .= n - 2
    .= n -' 2 by A5,XREAL_1:233;
  then
  ((2*n -' 3) choose (n -' 1)) = ((2*n -' 3)!) / ((n -' 1)! * ((n -' 2)!))
  by A2,NEWTON:def 3;
  then
A8: 4 * ((2*n -' 3) choose (n -' 1)) = (4 * ((2*n -' 3)!)) / ((n -' 1)! * ((
  n -' 2)!)) by XCMPLX_1:74;
A9: n -' 2 + 1 = n -' 1 by A1,Th4;
A10: n -' 1 = n - 1 by A1,XREAL_1:233;
  then
A11: n -' 1 + 1 = n;
A12: 1 * n < 2 * n by A1,XREAL_1:68;
  then
A13: 2 * n -' 1 = 2 * n - 1 by A1,XREAL_1:233,XXREAL_0:2;
  1 < 2 * n by A1,A12,XXREAL_0:2;
  then
A14: 2 * n -' 2 + 1 = 2 * n -' 1 by Th4;
  1 < 2 * n by A1,A12,XXREAL_0:2;
  then
A15: n -' 1 < 2 * n -' 1 by A12,NAT_D:57;
  2 * n -' 1 - (n -' 1) = 2 * n -' 1 - (n - 1) by A1,XREAL_1:233
    .= 2 * n -' 1 - n + 1
    .= 2 * n - 1 - n + 1 by A1,A12,XREAL_1:233,XXREAL_0:2
    .= n;
  then
  (2*n -' 1) choose (n -' 1) = ((2*n -' 1)!) / ((n -' 1)! * (n!)) by A15,
NEWTON:def 3;
  then
A16: (2*n -' 1) choose (n -' 1) = ((2*n -' 2)! * (2*n -' 1)) / ((n -' 1)! *
  (n!)) by A14,NEWTON:15;
  n - 1 > 0 by A1,XREAL_1:50;
  then
  4 * ((2*n -' 3) choose (n -' 1)) = ((n * (n - 1)) * (4 * ((2*n -' 3)!))
  ) / (((n * (n - 1)) * ((n -' 1)! * ((n -' 2)!)))) by A1,A8,XCMPLX_1:6,91
    .= (((n - 1) * n) * (4 * ((2*n -' 3)!))) / (((n - 1) * (n * ((n -' 1)!))
  * ((n -' 2)!)))
    .= (((n - 1) * n) * (4 * ((2*n -' 3)!))) / (((n - 1) * (n!) * ((n -' 2)!
  ))) by A11,NEWTON:15
    .= (((n - 1) * n) * (4 * ((2*n -' 3)!))) / (((n!) * ((n -' 1) * ((n -' 2
  )!)))) by A10
    .= (2 * n * ((2 * n -' 2) * ((2*n -' 3)!))) / (((n!) * ((n -' 1)!))) by A4
,A9,NEWTON:15
    .= (2 * n * ((2*n -' 2)!)) / (((n!) * ((n -' 1)!))) by A7,NEWTON:15;
  then
  4 * ((2*n -' 3) choose (n -' 1)) - ((2*n -' 1) choose (n -' 1)) = ((2 *
  n * ((2*n -' 2)!)) - ((2*n -' 2)! * (2*n -' 1))) / (((n!) * ((n -' 1)!))) by
A16,XCMPLX_1:120
    .= Catalan n by A1,A13,Th8;
  hence thesis;
end;
