reserve a,b,n for Element of NAT;

theorem
  for n being Element of NAT holds Lucas(n + 3) * Lucas(n) = (Lucas(n +
  2))|^2 - (Lucas(n+1))|^2
proof
  defpred P[Nat] means Lucas($1 + 3) * Lucas($1) = (Lucas($1 + 2))|^2 - (Lucas
  ($1+1))|^2;
A1: for k being Nat st P[k] & P[k+1] holds P[k+2]
  proof
    let k be Nat;
    assume that
    P[k] and
    P[k+1];
    Lucas((k+2)+3)* Lucas(k+2)=Lucas((k+3)+2)*Lucas(k+2)
      .=(Lucas(k+3)+Lucas((k+3)+1))*Lucas(k+2) by Th12
      .=(Lucas(k+3)+Lucas((k+2)+2))*Lucas(k+2)
      .=(Lucas(k+3)+(Lucas(k+2)+Lucas((k+2)+1)))*Lucas(k+2) by Th12
      .=(Lucas(k+2)+Lucas(k+3))*(Lucas(k+2)+Lucas((k+2)+1))- Lucas(k+3)*
    Lucas(k+3)
      .=(Lucas(k+2)+Lucas(k+3))*Lucas((k+2)+2)-Lucas(k+3)*Lucas(k+3) by Th12
      .=(Lucas(k+2)+Lucas((k+2)+1))*Lucas(k+4)-Lucas(k+3)*Lucas(k+3)
      .=Lucas((k+2)+2)*Lucas(k+4)-Lucas(k+3)*Lucas(k+3) by Th12
      .=Lucas(k+4)*Lucas(k+4)-(Lucas(k+3))|^2 by WSIERP_1:1
      .=(Lucas((k+2)+2))|^2-(Lucas((k+2)+1))|^2 by WSIERP_1:1;
    hence thesis;
  end;
  Lucas(1 + 3) * Lucas(1) = 4*4 - 3*3 by Th11,Th16
    .= 4*4 - 3|^2 by WSIERP_1:1
    .= (Lucas(1 + 2))|^2 - (Lucas(1+1))|^2 by Th14,Th15,WSIERP_1:1;
  then
A2: P[1];
  (Lucas(0 + 2))|^2 - (Lucas(0+1))|^2 = 3*3 - (Lucas(1))|^2 by Th14,WSIERP_1:1
    .= 9-1*1 by Th11,WSIERP_1:1
    .= Lucas(0+3) * Lucas(0) by Th11,Th15;
  then
A3: P[0];
  for k being Nat holds P[k] from FIB_NUM:sch 1 (A3, A2, A1);
  hence thesis;
end;
