We know that we can decide membership of "Hf" by way of a deterministic Turing machine that first calculates "f"(|"x"|), then writes out a row of 0s of that length, and then uses this row of 0s as a "clock" or "counter" to simulate "M" for at most that many steps. At each step, the simulating machine needs to look through the definition of "M" to decide what the next action would be. It is safe to say that this takes at most "f"("m")3 operations, so