reserve a,b,n for Element of NAT;

theorem
  for n,m being Element of NAT holds 2 * Lucas(n + m) = Lucas(n) * Lucas
  (m) + 5 * Fib(n) * Fib(m)
proof
  defpred P[Nat] means for n holds 2* Lucas(n + $1) = Lucas(n)*Lucas($1) + 5*
  Fib(n)* Fib($1);
A1: now
    let k be Nat;
    assume that
A2: P[k] and
A3: P[k+1];
    thus P[k+2]
    proof
      let n;
A4:   2* Lucas(n + (k+1)) = Lucas(n)*Lucas(k+1) + 5*Fib(n)* Fib(k+1) by A3;
      2* Lucas(n+(k+2))=2*Lucas(n+k+2) .= 2*(Lucas(n+k)+Lucas((n+k)+1)) by Th12
        .= 2*Lucas(n+k)+2*Lucas(n+k+1)
        .= (Lucas(n)*Lucas(k) + 5*Fib(n)* Fib(k)) + 2*Lucas(n+(k+1)) by A2
        .= Lucas(n)*(Lucas(k) + Lucas(k+1))+5*Fib(n)* (Fib(k) + Fib(k+1)) by A4
        .= Lucas(n)*Lucas(k+2) + 5*Fib(n)*(Fib(k) + Fib(k+1)) by Th12
        .= Lucas(n)*Lucas(k+2) + 5*Fib(n)* Fib(k+2) by FIB_NUM2:24;
      hence thesis;
    end;
  end;
A5: P[1]
  proof
    let n be Element of NAT;
    2*Lucas(n + 1) = Lucas(n+1)+Lucas(n)+Lucas(n+1)-Lucas(n)
      .= Lucas(n+1)+Lucas(n+2)-Lucas(n) by Th12
      .= Lucas(n+3)-Lucas(n) by Th13
      .= Lucas(n) + (Lucas(n + 3) - 2* Lucas(n))
      .= Lucas(n)*Lucas(1) + 5*Fib(n)* Fib(1) by Th11,Th23,PRE_FF:1;
    hence thesis;
  end;
A6: P[0] by Th11,PRE_FF:1;
  for k being Nat holds P[k] from FIB_NUM:sch 1 (A6, A5, A1);
  hence thesis;
end;
