Is there a reduction from ATM to ATM-complement?

(ATM denotes the language $\{\langle M,w \rangle \mid \text{TM $M$ accepts $w$}\}$)

I have been thinking about it too much and couldn't find the answer.

I know that reduction from ATM-complement to ATM is not possible becouse if it was, ATM would not be in RE. But how can I proove/profe the other way around?


1 Answer 1


By the very same reason you have mentioned there cannot be such a (many-one) reduction.

Note that if you have two problems, say $A$ and $B$, and $A\le_m B$ then also $\bar A \le_m \bar B$. In fact the same reduction can be used for both statements. So a reduction from ATM to ATM-complement, would also be a reduction from ATM-complement to ATM. And you already know that such a thing does not exist.

If you are however ask for Turing reduction, then the answer is yes. We can ask an ATM oracle and invert the answer to get a Turing machine that decides ATM-complement.


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