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aa:beyondre [2016/10/31 14:25]
rstefan [A general undecidability result (Rice's Theorem)]
aa:beyondre [2016/10/31 14:31] (current)
rstefan
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 $math[\Pi_M(\omega) = \mbox{ if } M_x(M) \mbox{halts,​ then run} M^*(\omega) \mbox{.}] $math[\Pi_M(\omega) = \mbox{ if } M_x(M) \mbox{halts,​ then run} M^*(\omega) \mbox{.}]
  
-If $math[f_{\Pi_w}] is the problem accepted by $math[\Pi_w], we show that:+If $math[f_{\Pi_M}] is the problem accepted by $math[\Pi_M], we show that:
  
-$math[f_{\Pi_w} \in \mathcal{C} \mbox{ iff } M_x(w) \mbox{ halts}]+$math[f_{\Pi_M} \in \mathcal{C} \mbox{ iff } M_x(M) \mbox{ halts}]
  
-$math[(\Rightarrow)]. Suppose $math[f_{\Pi_w} \in \mathcal{C}]. Then $math[\Pi_w(\omega)] cannot loop for every input $math[\omega \in \Sigma^*]. If there were so, then $math[f_{\Pi_w}] would be the trivial function always returning $math[0] for any input, which we have assumed is not in $math[\mathcal{C}]. Thus, $math[M_x(w)] halts.+$math[(\Rightarrow)]. Suppose $math[f_{\Pi_M} \in \mathcal{C}]. Then $math[\Pi_M(\omega)] cannot loop for every input $math[\omega \in \Sigma^*]. If there were so, then $math[f_{\Pi_M}] would be the trivial function always returning $math[0] for any input, which we have assumed is not in $math[\mathcal{C}]. Thus, $math[M_x(M)] halts.
  
-$math[(\Leftarrow)]. Suppose $math[M_x(w)] halts. Then the +$math[(\Leftarrow)]. Suppose $math[M_x(M)] halts. Then the 
- ​behaviour of $math[\Pi_w(\omega)] is precisely that of + ​behaviour of $math[\Pi_M(\omega)] is precisely that of 
-$math[M^*(\omega)]. $math[\Pi_w(\omega)] will return $math[1] whenever $math[M^*(\omega)] will return $math[1] and $math[\Pi_w(\omega) = \perp] whenever $math[M^*(\omega) = \perp]. Since $math[f \in \mathcal{C}],​ then also $math[f_{\Pi_w} \in \mathcal{C}].+$math[M^*(\omega)]. $math[\Pi_M(\omega)] will return $math[1] whenever $math[M^*(\omega)] will return $math[1] and $math[\Pi_M(\omega) = \perp] whenever $math[M^*(\omega) = \perp]. Since $math[f \in \mathcal{C}],​ then also $math[f_{\Pi_M} \in \mathcal{C}].
 $end $end