Rice's theorem and the Halting problem


by Jaime Sevilla Molina Jul 29 2016 updated Aug 14 2016

We will show that Rice's theorem and the the halting problem are equivalent.

The Halting theorem implies Rice's theorem

Let $~$S$~$ be a non trivial set of computable partial functions, and suppose that there exists a Turing machine encoded by $~$[n]$~$ such that: $$~$ [n](m) = \left\{ \begin{array}{ll} 1 & [m] \text{ computes a function in $S$} \\ 0 & \text{otherwise} \\ \end{array} \right. $~$$

We can assume [without_loss_of_generality w.l.o.g.] that the empty function undefined on every input is not in $~$S$~$ %%note:Suppose that the empty function is in $~$S$~$. Then it is satisfied that the empty function is not in $~$S^c$~$, and if $~$S$~$ is decidable then it follows immediately that [thecomplementofadecidablesetis_decidable $~$S^c$~$ is decidable as well]. So we can use $~$S^c$~$ as our $~$S$~$ and the argument follows exactly the same.%%. Thus there exists a computable function in $~$S$~$ computed by some machine $~$[s]$~$ such that $~$[s](x)$~$ is defined for some input $~$x$~$.

Suppose we want to decide whether the machine $~$[m]$~$ halts on input $~$[x]$~$.

For that purpose we can build a machine $~$Proxy_s$~$ which does the following:

    call [m](x)
    return s[z]

Clearly, if $~$[m](x)$~$ halts then Proxy_z computes the same function as $~$[s]$~$, and thus $~$[n](Proxy_s)=1$~$.

If on the other hand $~$[m](x)$~$ does not halt, then Proxy_s(z) computes the empty function, which we assumed to not be in $~$S$~$, and therefore $~$[n](Proxy_s)=0$~$.

Thus we can use a Turing machine computing pertinence to $~$S$~$ to decide the halting problem, which we know is undecidable. We conclude that such a machine cannot possibly exists, and thus Rice's theorem holds.

Rice's Theorem implies the Halting theorem

Suppose that there was a Turing machine $~$HALT$~$ deciding the Halting Problem.

Let $~$S$~$ be the set of computable functions defined on a fixed input $~$x$~$, which is clearly non-trivial, as it does not contain the empty function but is not empty either. Let $~$[n]$~$ be a Turing machine, and we want to decide whether $~$[n]\in S$~$ or not. If this was possible for an arbitrary $~$[n]$~$, then we would have reached a contradiction, as Rice's theorem forbids this outcome.

But $~$[n]$~$ belongs to $~$S$~$ iff $~$[n]$~$ halts on input $~$x$~$, so we can use $~$HALT$~$ to decide whether $~$[n]$~$ belongs to $~$S$~$, in contradiction with Rice's theorem. So our supposition of the existence of $~$HALT$~$ was erroneous, and thus the Halting theorem is true.


Patrick Stevens

We will show that Rice's theorem and the halting problem are equivalent to each other\.

I think the halting problem probably should have its own page, rather than being linked to the umbrella uncomputability page.

Patrick Stevens

Surely they are equivalent. Given a Rice-deciding oracle, we can ask the oracle, "Does the partial function defined by machine $~$[n]$~$ specify where input $~$k$~$ should go?"; that determines whether $~$[n]$~$ halts on input $~$k$~$ or not.