Undecidability of: $|w in L| geq 1, L={w in {0,1}^*|a_0·#_0(w)+a_1·#_1(w)- a_1a_0=0}$












2












$begingroup$


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?






turing-machines decidability







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















2












$begingroup$


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?






turing-machines decidability







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












2








2





$begingroup$


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?






turing-machines decidability







share|cite|improve this question











$endgroup$




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?






turing-machines decidability




turing-machines decidability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 10:14







nocturne

















asked Nov 30 '18 at 10:11









nocturnenocturne

687




687












  • $begingroup$
    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




















  • $begingroup$
    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


















$begingroup$
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






$begingroup$
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 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




$begingroup$
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






$begingroup$
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












1 Answer
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$begingroup$

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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    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 10:30









        plattyplatty

        3,370320




        3,370320






























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