For example, because the decidable languages are closed under union … 3 I'm having a problem with proving the following statement: For every infinite language $L$, does there exists an infinite language $L' \subseteq L$ such that $L'$ is not … The source from where I got this question is telling that the answer is undecidablebut I think it is decidable Because : since sigma * is infinite (correct me if I am … I found some examples of infinite regular languages having non-regular subsets. Show language and its complement are both recognizable: If is T-recog and is T-recog then is … How would one go about proving/disproving the language { A ∣A is an NFA and L (A)= {0,1}∗} is/isn't decidable? I assumed at first since it was an NFA involved it would be … I came across these 2 problems about proving of undecidability of languages: $1$. A decider that recognizes language … Decidability is making decision about following: Is given FA/RE accept any string or not? Is given FA/RE represents finite or infinite language? Whether two languages are equivalent or not? Every TM for a semi-decidable+ language halts in the accept state for strings in the language but loops for some strings not in the language. If it doesn't, Since M has finite number … An algorithm or recipe for recognizing languages is anything that can be fed a word over the given alphabet, and that depending on its input either "accepts" the input after some time, or … We will identify which of the 24 problems are decidable and, if they do not appear decidable, prove that they are undecidable. thanks, I know that's a weaker statement, but it's bold that there could be finite and infinite Undecidable Languages, and I think this special case must be included in your answer, the … The language is infinite iff its grammar can generate an infinite number of words, or equivalently iff a recognizing automaton can recognize an infinite number of words. 1 shows each problem’s decidability property. Does this work? The above was an example of a language, didn't think it would be recognizable. $FINITE_ {\text {TM}} = \ {\langle M \rangle | M \text { is a Turing machine and } L Nope, same idea, take the empty language, which is contained in every language, but is decidable. For example, … If I have the language $\ {a\}^*$ for example, that's infinite but you can make a DFA for any sub-language of it, right? There's a hint that this can be proved using diagonalization, … Since our goal in this section is to find an explicit undecidable language, it is natural to consider applying Lemma (Diagonalization). Dans le cas d’un langage rationnel, on a : — Accept est décidable, — Empty est décidable, — Univ est décidable, — Empty-Intersection est (in)décidable, — Egal est décidable. rhtzx
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