reserve a,b,n for Element of NAT;

theorem
  for n being Nat holds 2 * Lucas(n) + Lucas(n + 1) = 5 * Fib(n + 1)
proof
  defpred P[Nat] means 2*Lucas($1)+Lucas($1+1)=5*Fib($1+1);
A1: for k being Nat st P[k] & P[k+1] holds P[k+2]
  proof
    let k be Nat;
    assume that
A2: P[k] and
A3: P[k+1];
    2* Lucas(k+2) + Lucas((k+2)+1) = 2* Lucas(k+2) + Lucas(k+(2+1))
      .= 2* Lucas(k+2) + (Lucas(k+1) + Lucas(k+2)) by Th13
      .= 2* Lucas(k+2) + Lucas(k+1) + Lucas(k+2)
      .= 2* Lucas(k+2) + Lucas(k+1) + (Lucas(k) + Lucas(k+1)) by Th12
      .= Lucas(k+2) + (2*Lucas(k+1) + Lucas(k+2)) + Lucas(k)
      .= Lucas(k) + Lucas(k+1) + 5*Fib(k+2) + Lucas(k) by A3,Th12
      .= 5*Fib(k+1) + 5*Fib(k+2) by A2
      .= 5*(Fib(k+1)+Fib(k+(1+1)))
      .= 5*Fib(((k+1)+1)+1) by PRE_FF:1
      .= 5*Fib((k+2)+1);
    hence thesis;
  end;
  0+1+1=2;
  then Fib(2) = 1 by PRE_FF:1;
  then
A4: P[1] by Th11,Th14;
A5: P[0] by Th11,PRE_FF:1;
  thus for k being Nat holds P[k] from FIB_NUM:sch 1 (A5,A4,A1);
end;
