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Let $a_0, a_1 \in \mathbb{N} \setminus \{0\}$ and $L=\{w \in \{0,1\}^*|a_0·\#_0(w)+a_1·\#_1(w)- a_1a_0=0\}$ . Let's assume problem $P$ that, language of Turing machine accepts at least one word from language $L$. Using membership problem, is $P$ undecidable?

Membership problem of $w \in L \land L \in unrestricted$, is undecidable but is semi-decidable. Prove of semi-decitability is quite easy by simulating of Turing machine $T$, where $L(T)=L$. But how can we show that $P$ is undecidable?

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  • $\begingroup$ You don't include a condition in the set-builder notation. What is this supposed to say? It's also unclear what you mean by $L \land L$. $\endgroup$ – platty Nov 30 '18 at 10:12
  • $\begingroup$ I edited the question, is it better? $\endgroup$ – nocturne Nov 30 '18 at 10:16
  • $\begingroup$ I think so. Is $P$ supposed to be a language of Turing Machine encodings, and the question is whether $P$ is undecidable? $\endgroup$ – platty Nov 30 '18 at 10:16
  • $\begingroup$ Question is whether the P is undecidable. Problem P is undecidable when is not decidable which means we can construct a Turing machine $T$, where it decides whether it accepts $w \in L$ and rejects $w \in \sum^* \setminus L$ $\endgroup$ – nocturne Nov 30 '18 at 10:22
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To show that a language over Turing Machine encodings is undecidable, we can use Rice's Theorem, which states that any non-trivial semantic property of Turing Machines is undecidable. Here, our property is "Turing Machine $M$ accepts at least one word from language $L$." We first note that, for every $a_0,a_1 \in \mathbb{N} \setminus \{0\}$, $L$ is nonempty and does not contain every string. For the former, $1^{a_0} \in L$ for any $a_0,a_1$; for the latter, observe that $\varepsilon \notin L$ for any such $L$.

From here, we just need to show that this property is nontrivial. This requires showing that there is a Turing machine that accepts a language satisfying this property, and one that accepts a language which does not satisfy this property. Turing machines accepting the languages $\Sigma^*$ and $\emptyset$ suffice, respectively.

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