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This is a problem from Sipser's book (marked with an asterisk).

$EQ_{TM} = \{(\langle M \rangle, \langle N \rangle)$ where $M$ and $N$ are Turing machines and $L(M) = L(N)\}$

We know that neither $EQ_{TM}$ nor $\overline{EQ_{TM}}$ are recognizable so unsure how to go about proving there can't be a mapping reduction from one to the other.

Any hints?

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    $\begingroup$ Where exactly did you get stuck? What are your thoughts? $\endgroup$ – vonbrand Jan 29 '16 at 17:12
  • $\begingroup$ First I realized that it's not useful to use the reduction from any other language to either $EQ_{TM}$ or $\overline{EQ_{TM}}$ and then look for contradiction (the language would then be recognizable or decidable) - because we know they are not recognizable. $\endgroup$ – user1255841 Jan 29 '16 at 17:16
  • $\begingroup$ Then I tried to use the fact that I have a 1-to-1 map from $<M_1, M_2>$ to $<N_1,N_2>$ where $L(M_1) = L(M_2)$ and $L(N_1) \ne L(N_2)$ . Now I am trying to find a language that I can construct a TM to recognize using this map - and so far not finding any $\endgroup$ – user1255841 Jan 29 '16 at 17:19
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Your language is $\Pi_2$-complete: it can be written as a $\Pi_2$ formula, and TOT can be reduced to it. The complement is therefore $\Sigma_2$-complete. No computable reductions can exist between $\Pi_2$-complete and $\Sigma_2$-complete sets (why?).

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  • $\begingroup$ Can you clarify what Pi_2 - complete means as well as what TOT refers to and what Sigma_2 complete means? $\endgroup$ – NateSHolland Feb 6 '16 at 15:06
  • $\begingroup$ These are levels of the polynomial hierarchy, explained for example on Wikipedia. TOT is the language of all Turing machines that halt on all inputs. $\endgroup$ – Yuval Filmus Feb 6 '16 at 16:34
  • $\begingroup$ I see, and so the inequality is NP complete and checking equality is co-NP complete and we can't reduce between the two? $\endgroup$ – NateSHolland Feb 6 '16 at 16:37
  • $\begingroup$ It's not known to be in NP. In fact, it's conjectured to be in neither NP nor coNP. It's both NP-hard and coNP-hard, though. $\endgroup$ – Yuval Filmus Feb 6 '16 at 16:45
  • $\begingroup$ Which one is NP-hard and which one is CoNP-hard? $\endgroup$ – NateSHolland Feb 6 '16 at 16:52
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Suppose a computable mapping reduction $t: \Sigma^* \to \Sigma^*$ from $EQ_{TM}$ to $\overline{EQ_{TM}}$ exists.

We want to create TM's $A,B$ such that for $\langle A', B' \rangle = t(\langle A,B \rangle)$, we have \begin{align} L(A) &= L(A') \\ L(B) &= L(B') \end{align}

We will nest the recursion theorem to accomplish this goal. Our "inner" use of the recursion theorem will be explicit.

Let $F$ be the following TM, which ignores its input:

  1. Obtain own description $\langle F \rangle$ via the recursion theorem (Sipser 6.3)
  2. Write out the description $\langle A_0 \rangle$ of the following TM. On input $\langle M,w \rangle$:
    • Save the constant string $\langle F \rangle$
    • Save $\langle M \rangle$
    • Simulate step (4) and (5) of $\langle F \rangle$ to get $\langle B \rangle$
    • Compute $\langle M', X' \rangle = t(\langle M, B \rangle)$
    • Simulate $\langle M' \rangle$ on $w$.
  3. Use the recursion theorem to construct $A$ which computes $A_0(\langle A, w \rangle)$ on input $w$.
  4. Write out the description $\langle B_0 \rangle$ of the following TM. On input $\langle M,w \rangle$:
    • Save the constant string $\langle F \rangle$
    • Save $\langle M \rangle$
    • Simulate step (2) and (3) of $\langle F \rangle$ to get $\langle A \rangle B$
    • Compute $\langle X', M' \rangle = t(\langle A, M \rangle)$.
    • Simulate $\langle M' \rangle$ on $w$.
  5. Use the recursion theorem to construct $B$ which computes $B_0(\langle B, w \rangle)$ on input $w$
  6. print $\langle A, B \rangle$.

Note that when running the TM $A_0$ (and analogously for $B_0$), the simulation of $\langle F \rangle$ is purely "code-manipulation" of the program $\langle B_0 \rangle$: both writing out the description and constructing the recursive variant do not involve running $B_0$ at all. The latter is apparent from a close reading of the definition of $\langle R \rangle$ in Sipser, page 249. This ensures the definitions of $A$ and $B$ are not circular.

Running $F$ generates $\langle A,B \rangle$. Let $\langle A', B' \rangle = t(\langle A,B \rangle)$. Since $A$ simulates $A'$ and $B$ simulates $B'$, we have that $L(A) = L(A')$ and $L(B) = L(B')$.

It must be the case that $L(A) = L(B)$ or $L(A) \neq L(B)$. In the first case, we have \begin{align*} L(A) = L(B) = L(B') \neq L(A') = L(A) \end{align*} In the second case, \begin{align*} L(A) \neq L(B) = L(B') = L(A') = L(A) \end{align*} In either case we have elicited a contradiction of the assumption that the mapping $t$ exists.

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If there were a reduction $f$ from $EQ_\mathsf{TM}$ to $\overline{EQ_\mathsf{TM}}$, then it would be a computable function satisfying that

$$f(\langle M_1,M_2 \rangle) = \langle M'_1, M'_2 \rangle$$

where $L(M_1) = L(M_2) \iff L(M'_1) \neq L(M'_2)$.

We would then also have computable functions $f_1, f_2 : \Sigma^* \rightarrow \Sigma$ defined by

$$f_1((\langle M_1 \rangle, \langle M_2 \rangle)) = \langle M'_1 \rangle \text{ if } f(\langle M_1,M_2 \rangle) = \langle M'_1, M'_2 \rangle$$

$$f_2(\langle M_1 \rangle, \langle M_2 \rangle)) = \langle M'_2 \rangle \text{ if } f(\langle M_1,M_2 \rangle) = \langle M'_1, M'_2 \rangle$$

But then by the first recursion theorem, we would have "curried" versions of these for any given $M$:

$$f^M_1(\langle M_2 \rangle) = \langle M'_1 \rangle \text{ if } f(\langle M,M_2 \rangle) = \langle M'_1, M'_2 \rangle$$

$$f^M_2(\langle M_1 \rangle) = \langle M'_2 \rangle \text{ if } f(\langle M_1,M_2 \rangle) = \langle M'_1, M'_2 \rangle$$

By the fixed-point theorem, there must then be a machine $F$ such that

$$f^M_1(\langle F \rangle) = \langle F' \rangle$$

where $L(F') = L(F)$ and a machine $G$ such that

$$f^M_2(\langle G \rangle) = \langle G' \rangle$$

and $L(G') = L(G)$.

But then we get that $f(\langle F, G \rangle) = \langle f^F_1(\langle G \rangle), f^G_2(\langle F \rangle) \rangle = (\langle F', G' \rangle)$. Clearly we do not have that $L(F) = L(G) \iff L(F') \neq L(G')$.

We have thus reached a contradiction, using only the assumption that $f$ existed.

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  • $\begingroup$ I think this proof is incorrect. In the equation $\langle f_1^F(G), f_2^G(F) \rangle = \langle F^\prime, G^\prime \rangle$, the fixed-point theorem only handles unary functions and thus does not guarantee existence of $F^\prime$ and $G^\prime$ with the same languages as their counterparts on the left-hand side. The fixed points $F$ and $G$ are defined here in terms of each other (in particular, we have $M = G$ when finding the fixed point of $f_1^M$, but $M = F$ for $f_2^M$) and thus there is a circular definition that cannot be computable. $\endgroup$ – xdavidliu Oct 20 at 16:46
  • $\begingroup$ Lambda notation makes this easier to express. To make this more readable, I do not use the angular brackets for pairs. Define the curried version of the reduction f by $$f_C = \lambda X. \lambda Y. f(X,Y)$$ Then we let $f^1 = f_C(F)$ and $f^2 = f_C(G)$. $f^1$ has a fixed point, $F’$. $f^2$ has a fixed point, $G’$. This follows from Rogers’ fixed point theorem and the assumption that f is total and recursive. Clearly, $f(F,G) = (f^1(G),f^2(G))$. But now $f(F’,G’) = (F’,G’)$. And then we have, by the faithfulness assumption for $f$ that $L(F’) = F(G’)$ iff $L(F’) \neq L(G’)$. $\endgroup$ – Hans Hüttel Oct 20 at 18:49
  • $\begingroup$ Some of the results in your comment don't make sense. If you we take the definition of $f^1$ and $f^2$, we have $f^1 = \lambda Y . f(F, Y)$, and thus $f^1(G) = f(F, G)$. Similarly, $f^2(G) = f(G, G)$. Then it is not true that $f(F, G) = (f^1(G), f^2(G))$. The comment doesn't address my main point that you cannot compute the fixed points here, because they depend on each other: in your answer $f_1^M(F) = F^\prime$, this entire equation is parameterized by $M$. You cannot use the fixed point theorem to compute both $F$ and $G$, since each computation depends on the result of the other. $\endgroup$ – xdavidliu Oct 20 at 19:30
  • $\begingroup$ In other words, this answer says that $G$ is the fixed point of $f_1^F$ while $F$ is the fixed point of $f_2^G$. This definition clearly has a cycle, and thus cannot be computed using the fixed point theorem. $\endgroup$ – xdavidliu Oct 20 at 19:48
  • $\begingroup$ Note that it may be tempting to argue something like this: "let $N$ be the fixed point of $f_1^M$ and let $Q$ be the fixed point of $f_2^N$. This is a computable function $g$ taking $M$ to $Q$ and thus has a fixed point $F$." However the problem with this argument is that $g$ may not necessarily take $F$ to itself; it only needs to take $F$ to a possibly different machine with the same language. Hence, this construction cannot be used to resolve the circular definition. $\endgroup$ – xdavidliu Oct 21 at 3:34
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Suppose $EQ_\mathrm{TM} \leq_\mathrm{m} \overline{EQ_\mathrm{TM}}$. Then there exists an $f : \Sigma^\star \rightarrow \Sigma^\star$ such that $f(w) \in EQ_\mathrm{TM}$ if and only if $w \in \overline{EQ_\mathrm{TM}}$.

Without loss of generality, we can assume $f(w)$ is always a pair $\langle M_1, M_2 \rangle$ of TMs, as opposed to gibberish in the case that $w \in EQ_\mathrm{TM}$. To ensure this, wrap $f$ with a simple procedure that converts any $f(w)$ that is not a pair of TMs to some (doesn't matter which) pair of non-equivalent TMs.

Now for any TMs $A$ and $B$, we have $f(A, B) = \langle A^\prime, B^\prime \rangle$, where $L(A^\prime) \neq L(B^\prime)$ if and only if $L(A) = L(B)$.

Now define the function $a(B)$ as the $A$ that, given $B$, will result in $L(A^\prime) = L(A) = L(a(B))$ above. This function $a(B)$ is computable using the fixed point version of the recursion theorem (6.8 in Sipser) with input $A$ and output $A^\prime$.

Now let $p(B)$ be the $B^\prime$ that results from $f(a(B), B)$. Let $P$ be the fixed point of $p$, i.e. $L(P) = L(p(P))$. We then have $f(a(P), P) = \langle A^\prime, p(P)\rangle$. By construction, we still have $L(a(P)) = L(A^\prime)$.

Can we have $L(a(P)) = L(P)$? No, because if so, then from the preceding equalities, we would also have $L(a(P)) = L(A^\prime) = L(p(P))$, and thus $f$ would be taking an element of $EQ_\mathrm{TM}$ to another element of $EQ_\mathrm{TM}$, which it is forbidden from doing by definition. Similarly, we cannot have $L(a(P)) \neq L(P)$, since that would result in $L(a(P)) = L(A^\prime) \neq L(p(P))$ and thus $f$ would be taking an element of $\overline{EQ_\mathrm{TM}}$ to another element $\overline{EQ_\mathrm{TM}}$, which it is forbidden from doing.

Since we have ruled out all possibilities, we see that $EQ_\mathrm{TM} \nleq_\mathrm{m} \overline{EQ_\mathrm{TM}}$.

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