# Undecidability

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L1:{<M> | there exist a Turing machine M' such that <M>$\neq$<M'> and L(M) = L(M')}

How this problem becomes trivial? and if it non-trivial then please explain why is that so. According to my understanding, non-trivial properties are the one where a language or string may get accepted by some Turing machine and not by some other Turing machines and hence it becomes undecidable. Hence it becomes undecidable when a given Turing machine it will accept a given string or not as it becomes non-trivial property.

For every non deterministic TM M1 there exists an equivalent deterministic TM M2 recognizing the same language, in this case we will have 2 different machines and both will accept same language, is this description holds true?? and if it is then is it okay to say M1=M2 because they are kind of same machine but other one is just with some non deterministic nature.
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I think the language is decidable ..for a language to be  decidable we have to answer a question for every string  that YES it belongs to L or NO it does not belongs to L

Given any Valid TM encoding ..we have to answer whether there is one TM different from this and accept same language.We know for every NTM <---> DTM which accepts same language hence Our question is trivial with answer YES always ..Hence we can say that language is decidable..

correct me if wrong

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Use Rice’s theorem, to prove the undecidability of each of the following languages. $INFINITE_{TM} = \{\langle M \rangle \mid \text{M is a TM and L(M) is an infinite language}\}$. $\{\langle M \rangle \mid \text{M is a TM and }\:1011 \in L(M)\}$. $ALL_{TM} = \{\langle M \rangle \mid \text{ M is a TM and}\: L(M) = Σ^{\ast} \}$.
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