Decidability of the TM's computing a non-empty subset of total functions

Question

I have this HW problem: Let $F$ be the set of computable total functions, and let $\emptyset\subsetneq S\subseteq F$. Denote $$L_S=\{ \langle M \rangle | M \text{ is a TM that computes a function and } f_M\in S \}$$

where $f_M$ is the function $M$ computes.

Prove that for every such none-trivial $S$, $L_S \notin \mathcal{R}$.

I tried to construct

• $L_{f}=\left\{ x\#y\in{\Sigma^*}|y=f\left(x\right)\right\} $
• $\tilde{S}=\left\{ L_{f}|f\in S\right\} $
• $L_{\tilde{S}}=\left\{ \left\langle M\right\rangle |L\left\langle M\right\rangle \in\tilde{S}\right\} $

and then show with Rice that $L_{\tilde{S}} \notin \mathcal{R}$, when the idea behind it was to eventually show that $L_{\tilde{S}} \leq_m L_S $. But the problem here is that I couldn't show a mapping reduction from $L_{\tilde{S}}$ to $L_S$ without assuming $L_{\tilde{S}} \in \mathcal{RE}$ (which I'm quite sure is not true).
So any other directions will be warmly welcomed!

Answer 1

For any function $f: \Sigma^* \rightarrow \Sigma^*$, let

$$L_f = \{ x \# y \mid y = f(x) \}$$

For any Turing machine $M$, we can construct the description of a Turing machine $M'$ such that $M'$ recognizes $L_f$ if and only if $M$ computes $f$. $M'$ will receive a string, check that it is of the form $x \# y$ and then simulate $M$ with input $x$. If the simulation returns $w$, $M'$ accepts if $w = y$. Therefore, the language

$$ A_f = \{ \langle M \rangle \mid M \text{ is a TM that computes the function $f$} \}$$

reduces to the language

$$B_f = \{\langle M \rangle \mid L(M) = L_f \}$$

The class

$$L_S = \{ L_f \mid f \in S \}$$

is a nontrivial class of Turing-recognizable languages and therefore the decision problem $B_f$ is undedicable by Rice's theorem.

Answer 2

Assume WLOG* $Id\in S$ ($\forall x\in\Sigma^*:\ Id\left(x\right)=x$)

Define $f\left(\left\langle M,w\right\rangle \right)\mapsto\left\langle M'\right\rangle$ where $M'$ is a TM that on input x:

• if $M$ halts on $w$ then $M'$ accept (leave $x$ on the tape)

if $\left\langle M,w\right\rangle \in H_{TM}$ then $M'$ halts on every $x$ and the content of the tape by the end of the run is $x$ ($M'$ computes $Id$) - $\left\langle M'\right\rangle \in F$

if $\left\langle M,w\right\rangle \notin H_{TM}$ then $M'$ never halts hence $\left\langle M'\right\rangle \notin F$.

Since $H_{TM}\leq_m L_{S}$ and $H_{TM}\notin\mathcal{R}$ we have $L_{S}\notin\mathcal{R}$.

WLOG*: if $Id\notin S$ prove for $\overline{S}=F\smallsetminus S$ - since now $Id\in\overline{S}$ the same proof holds, so $L_{\overline{S}}\notin\mathcal{R} L_{\overline{S}}=\left\{ \left\langle M\right\rangle |f_{M}\notin S\right\}$ . Assume that $\overline{L_{S}}=\left\{ w\in\Sigma^*|\mbox{ if }w\mbox{ is an encoding of a TM then }w\mbox{ does not compute a function }f\in S\right\} \in\mathcal{R}$ and $R$ decides $\overline{L_{S}}$. We can construct a reduction f that on input $w$:

• if $R$ accepts $w$

  • check if w is a TM encoding
  • if w is a TM, accept

• reject

then $w\in\overline{L_{S}}$ iff $f\left(w\right)\in L_{\overline{S}}$ so we can decide $L_{\overline{S}}$ using $R$.

So $\overline{L_{S}}\notin\mathcal{R}\implies L_{S}\notin\mathcal{R}$

Answer 3

You can construct $\ \mathcal{C}=\{L\in\mathcal{RE}\ |\ \exists f\in S \ \text{such that}\ \ \mathrm{Im} f=L\}$ and it is clear That $\emptyset \subset \mathcal{C}\subset \mathcal{RE}$ (because it is a set of languages from $\mathcal{RE}$ and it is clearly not empty and there are languges in $\mathcal{RE}$ that do not compute a computable function because they are not decidable) and then by Rice's Theorem you get $\{\langle M\rangle \ | \ L(M)\in\mathcal{C}\}\not \in \mathcal{R}$ and this language is exactly $L_S$ so $L_S \not \in \mathcal{R}$.