
theorem
  for n, A, B, N being Nat st Fusc N = A * Fusc (2*n+1) + B * Fusc (2*n+1+1)
  holds Fusc N = A * Fusc n + (B+A) * Fusc (n+1)
proof
  let n, A, B, N be Nat such that
A1: Fusc N = A * Fusc (2*n+1) + B * Fusc (2*n+1+1);
  2*n+1+1 = 2* (n + 1) & Fusc (2*n+1) = Fusc n+Fusc(n+1) by Th15;
  hence Fusc N = A*Fusc n+A*Fusc (n+1)+B*Fusc (n+1) by A1,Th15
    .= A * Fusc n + (B+A) * Fusc (n+1);
end;
