reserve a,b,n for Element of NAT;

theorem Th10:
  for n being Nat holds tau_bar to_power n + tau_bar to_power(n+1)
  = tau_bar to_power(n+2)
proof
  defpred P[Nat] means tau_bar to_power $1 + tau_bar to_power($1+1)= tau_bar
  to_power($1+2);
  let n be Nat;
A1: tau_bar to_power 0 + tau_bar to_power(0+1) =1+tau_bar to_power(1) by
POWER:24
    .=1+(1-sqrt 5)/2 by FIB_NUM:def 2,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_bar*tau_bar by FIB_NUM:def 2
    .=(tau_bar to_power 1)*tau_bar by POWER:25
    .=(tau_bar to_power 1)*(tau_bar to_power 1) by POWER:25
    .=(tau_bar)to_power(1+1) by FIB_NUM2:5
    .=tau_bar to_power(0+2);
A2: tau_bar+1= tau_bar + tau_bar to_power 0 by POWER:24
    .=tau_bar to_power(2) by A1,POWER:25;
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_bar to_power(k+2) + tau_bar to_power((k+2)+1) =tau_bar to_power(k+
    2)+(tau_bar to_power(k+2)*tau_bar to_power(1)) by FIB_NUM2:5
      .=tau_bar to_power(k+2)*(1+tau_bar to_power(1))
      .= tau_bar to_power(k+2)* tau_bar to_power(2) by A2,POWER:25
      .= tau_bar to_power((k+2)+2) by FIB_NUM2:5;
    hence thesis;
  end;
  tau_bar to_power 1 + tau_bar to_power(1+1) = tau_bar+tau_bar to_power (1
  +1) by POWER:25
    .=tau_bar+(tau_bar to_power 1)*(tau_bar to_power 1) by FIB_NUM2:5
    .=tau_bar+(tau_bar to_power 1)*(tau_bar) by POWER:25
    .=tau_bar+tau_bar*tau_bar by POWER:25
    .=tau_bar*(1+tau_bar)
    .=(tau_bar to_power(1))*(tau_bar to_power(2)) by A2,POWER:25
    .= tau_bar to_power(1+2) by FIB_NUM2:5;
  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;
