reserve a,b,n for Element of NAT;

theorem Th23:
  for n being Nat holds Lucas(n + 3) - 2 * Lucas(n) = 5 * Fib(n)
proof
  defpred P[Nat] means Lucas($1+3)-2*Lucas($1)=5*Fib($1);
A1: P[1] by Th11,Th16,PRE_FF:1;
A2: for k being Nat st P[k] & P[k+1] holds P[k+2]
  proof
    let k be Nat;
    assume that
A3: P[k] and
A4: P[k+1];
    Lucas(k+2+3) - 2*Lucas(k+2)= Lucas(((k+1+1+1)+1)+1) - 2* Lucas(k+2)
      .= Lucas(k+3)+Lucas(k+4) - 2* Lucas(k+2) by Th11;
    then Lucas(k+2+3) - 2*Lucas(k+2)= Lucas(k+3) + Lucas(k+4) - 2*(Lucas(k) +
    Lucas(k+1)) by Th12
      .= Lucas(k+4) - 2*Lucas(k+1)+ 5*Fib(k) by A3
      .= 5*Fib(k) + 5*Fib(k+1) by A4
      .= 5*(Fib(k) + Fib(k+1))
      .= 5*Fib(k+1+1) by PRE_FF:1
      .= 5*Fib(k+2);
    hence thesis;
  end;
A5: P[0] by Th11,Th15,PRE_FF:1;
  thus for k being Nat holds P[k] from FIB_NUM:sch 1 (A5,A1,A2);
end;
