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text: 'We will show that [5mv Rice's theorem] and the [46h the halting problem] are equivalent.\n\n#The Halting theorem implies Rice's theorem\nLet $S$ be a non trivial set of computable partial functions, and suppose that there exists a Turing machine encoded by $[n]$ such that:\n$$\n[n](m) = \n\\left\\{\n \\begin{array}{ll}\n 1 & [m] \\text{ computes a function in $S$} \\\\\n 0 & \\text{otherwise} \\\\\n \\end{array}\n \\right.\n$$\n\nWe can assume [without_loss_of_generality w.l.o.g.] that the empty function undefined on every input is not in $S$ \n%%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 [the_complement_of_a_decidable_set_is_decidable $S^c$ is decidable as well]. So we can use $S^c$ as our $S$ and the argument follows exactly the same.%%. \nThus there exists a computable function in $S$ computed by some machine $[s]$ such that $[s](x)$ is defined for some input $x$.\n\nSuppose we want to decide whether the machine $[m]$ halts on input $[x]$.\n\nFor that purpose we can build a machine $Proxy_s$ which does the following:\n\n Proxy_s(z):\n call [m](x)\n return s[z]\n\nClearly, if $[m](x)$ halts then Proxy_z computes the same function as $[s]$, and thus $[n](Proxy_s)=1$.\n\nIf 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 $[n](Proxy_s)=0$.\n\nThus 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.\n\n\n#Rice's Theorem implies the Halting theorem\nSuppose that there was a Turing machine $HALT$ deciding the Halting Problem.\n\nLet $S$ 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.\n\nBut $[n]$ 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.',
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