reserve n, k, r, m, i, j for Nat;

theorem Th25:
  for n being Nat holds Fib (n + 3) = Fib (n + 2) + Fib (n + 1)
proof
  let n be Nat;
  Fib (n+3) = Fib ((n+1)+2) .= Fib (n+1+1) + Fib (n+1) by Th24
    .= Fib (n+2) + Fib (n+1);
  hence thesis;
end;
