reserve a,b,n for Element of NAT;

theorem Th9:
  for n being Nat holds tau to_power n + tau to_power(n+1) = tau to_power(n+2)
proof
  defpred P[Nat] means tau to_power $1 + tau to_power($1+1)= tau to_power($1+2
  );
  let n be Nat;
A1: tau to_power 0 + tau to_power(0+1) =1+tau to_power(1) by POWER:24
    .=1+(1+sqrt 5)/2 by FIB_NUM:def 1,POWER:25
    .=(1+sqrt 5+ sqrt 5 + 5)/4
    .=(1+sqrt 5+ sqrt 5 + sqrt(5^2))/4 by SQUARE_1:22
    .=(1+1*sqrt 5+(sqrt 5)*1+(sqrt 5)*(sqrt 5))/4 by SQUARE_1:29
    .=tau*tau by FIB_NUM:def 1
    .=(tau to_power 1)*(tau) by POWER:25
    .=(tau to_power 1)*(tau to_power 1) by POWER:25
    .=(tau)to_power(1+1) by POWER:27
    .=tau to_power(0+2);
A2: 1 + tau = 1 + tau to_power 1 by POWER:25
    .= tau to_power(0+2) by A1,POWER:24;
A3: 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];
    tau to_power(k+2) + tau to_power((k+2)+1) =tau to_power(k+2)+(tau
    to_power(k+2)*tau to_power(1)) by POWER:27
      .=tau to_power(k+2)*(1+tau to_power(1))
      .= tau to_power(k+2)* tau to_power(2) by A2,POWER:25
      .= tau to_power((k+2)+2) by POWER:27;
    hence thesis;
  end;
  tau to_power 1 + tau to_power(1+1) =tau + (tau)to_power(1+1) by POWER:25
    .=tau + (tau to_power 1)*(tau to_power 1) by POWER:27
    .=tau + (tau to_power 1)*(tau) by POWER:25
    .=tau + (tau)*(tau) by POWER:25
    .=tau*(1+tau)
    .=(tau to_power(1))*(tau to_power(2)) by A2,POWER:25
    .= tau to_power(1+2) by POWER:27;
  then
A4: P[1];
A5: P[0] by A1;
  for k being Nat holds P[k] from FIB_NUM:sch 1 (A5, A4, A3);
  hence thesis;
end;
