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how to prove a language is not decidable Language, Logic and Intuition: A brief summary of the philosophy underpinning this site Logic and Language Why logic is not independent of language Formal Language: What a formal language is and what can be said within in and what can be said about it Natural Language and Reality: How philosophers have managed to become befuddled by natural language Mathematical Proofs 3. This states that a language is decidable if some Turing machine decides it. Deﬁnition: A language is co-enumerableif it is the complement of an enumerable language. Prove that B = B i BB B and " 2B. Sofya Raskhodnikova; based on slides by Nick Hopper Oct 13, 2011 · And again, ‘decidability’ comes down to the ability to recognize a valid proof, which rests on our innate axiomatic psychology, which math and logic have in common with language. Finally, the halting problem is undecidable, Proving that a language L is undecidable by reduction requires several steps. To see that, consider the following grammar: S! aSa 8a2 T S! a 8a2 T S! which generates the languages of palindromes. Decidability for a theory concerns whether there is an effective procedure that decides whether the formula is a member of the theory or not, given an arbitrary formula in the signature of the theory. sub. 20 Nov 2019 yes instances to P is not decidable or a language is undecidable if it is not very quick in coming up with proofs for these problems on the spot. In other words, prove that L = {h G i| G is a CFG over Σ = {0, 1} and 1? ∩ L (G) 6 = ∅} is decidable. Is a number 'm' prime? Solution. typeImpossible= (NotEvident:: Predicatek) Source#. Simulate B on w. The accepting states of Mare made nonaccepting states of Mwith no transitions, i. I (The proof is shown in the next slide). Mar 05, 2019 · Unrecognizable Language. [15 points] Solution:If A is decidable by some TM M, the enumerator operates by generating the strings in lexicographic order, testing each in turn for membership in A using M, and printing the string if it is in A. 16 (a) is solved in your book. A TM decides a language L iff it accepts all strings in L and rejects all strings not in L. The atomic formulas over this language are equality over string terms (word equations), linear inequality over length function (length constraints By a cardinality argument, we know that almost all languages are undecidable. The concatenation of languages K and L is the language KL = {xy|x ∈ K and y ∈ L}. . . Solution. Decidability of a theory. Question: Use Reduction To Prove That The Language Is Not Decidable ( And Are Turing Machines). de·cid·ed , de·cid·ing , de·cides v. L 4 = fh M ij is a TM that accepts all even numbers g. Use the pumping lemma to prove that Dec 01, 2010 · One can prove that such languages form a positive variety of languages. TM may halt in a rejecting configuration or go on indefinitely if w is not in the language. We construct a TM M0that decides the union of L 1 and L 2: M0 = \On input string w: 1. This decider ﬁrst converts a regular M accepts a language when it halts on every member string. Presumably the result of part (b) would be useful here. string w } A. The Halting Problem is an example of this third group. Given a Turing machine M one can construct a context-free grammar G such that L(G) is the set of invalid computations of M starting on blank tape. Proof. Lecture 4. If it could be decided, we could easily decide also the universal language Theorem 5. TM halts in an accepting If does not generate the algorithm doesn't halt. Sketch proofs that the class of Turing-decidable languages is closed under the language oper-ations union, intersection, complement, concatenation, and star. Jul 23, 2019 · These are also known as decidable languages. Write Regular Expression for the language that have the set of all. To choose not to accept someone. Reformulating Theorem: A language is decidable if and only if it is both We will prove that the languages are closed by creating the appropriate grammars. If Maccepts, reject. Proof in two directions: First, if A is decidable, show both A and its complement are Turing-recognizable. Proof: we know that HALT is CE but not decidable if complement of HALT wereCE, then HALT is CE and co-CE hence decidable. w. Reduce L2 to L1. •Use diagonalization in a proof of uncountability. Partially decidable or Semi-Decidable Language-– A decision problem P is said to be semi-decidable (i. Theirmotivation,aswithour work, is that veri cation of software requires decision procedures for expressive assertion languages. Thus, there exists a PDA P PAL that recognizes the language of palindromes. We know X does not exist. Brie y justify your answer for each statement. Languages decided by a TM are called decidable. The always-false property, which contains no RE languages. e. – Then we built Turing Machines H, H', D. Suppose that a language is acceptable but not decidable. 16 May 2018 We will present two proofs of existence of undecidable languages. Recursively Enumerable are accepted (recognised) by Turing Machines. Consider the function f(w) = 0w. I'll tell you the ingredients you need for the proof, but you'll have to figure out the recipe: - regular language 2. 3 is not decidable. Suppose now that BB B and " 2B. (c) Closed under complement. A decision problem P is called “undecidable” if the language L of all yes instances to P is not decidable. Languages recognized by a TM are called recognizable. PRELIMINARIES. For each of the following languages, state whether each language is (I) decidable, (II) semi-decidable but not decidable, or (III) not semi-decidable. denotes proper subsets. Turing-decidable languages. Therefore the resulting string is going to contain to have either num(1) num(2) or num(3) num(4) and is therefore not in the language C so the language C is not context free. (d) Closed under show that PALis decidable. Run M 1 on w. 1 axioms E. We provethat a set A ⊆ N is decidable if and only if A ×{0} and Ac ×{0} weakly self-assemble. Definition:A language is called semi-decidable(or recognizable) if there exists an algorithm that accepts a given string if and only if the string belongs to that language. , X = a TM that can decide ATM ) X 1. •Give examples of decidable problems. Assume Y exists. Consider the set of strings on {0,1} in which, every substring of 3 symbols has at most two zeros. } Divide the number ‘m’ by all the numbers between ‘2’ and ‘√m’ starting from ‘2’. This can be shown by reducing the halting problem to L: For the halting problem instance ( N, y ), create a new machine M for the L problem. in the book, or 𝑆𝑆. Empti-ness and ﬁniteness of tree automata with disequality and equality constraints between brothers (direct children) is proved decidable in [4]. 3 Check-in 8. 𝑇𝑇𝑇𝑇. Lecture 17: Proving Undecidability 5. Proving the Halting problem is undecidable. Solution: Let L= L(M) for some Turing Machine Mthat always halts. Prove that it is decidable whether or not L contains some string of the language 1?. (e. Class 1: those that have an algorithm (TM that halts whether or not it accepts Prove undecidable the language, Lu, consisting of pairs (M, w) such that. They use their decision procedure to prove that various CFG is a decidable language (see Lecture 3 or Theorem 4. L 2 = fh M ij is a TM and j 3 g. In fact, the following proof effectively shows that the complement of a context-free language may not even be recursive (i. bar] [summation over (term)] * such that L([X. Informally, is decidable i there is an algorithm which, given any sentence A, determines whether Ais in . If a theory T is complete (in the second sense, plus a second "technical" condition : effectively axiomatized), i. Show that if A <= m B and B is semi-decidable, then A is semi-decidable. So far, all of the languages presented above are recognizable, even though some of them are not decidable. If B = B, then BB B = B and " 2B = B. Prove your answer. T decides a language L if T recognizes L, and halts in all inputs. I By contraposition, if A is not decidable, then B is not decidable. To prove that a given language is Turing-recognizable: Construct an algorithm that accepts exactly those strings that are in the language. ﬂ So, to show a language L is not recursive using Rice’s Theorem, I pick the appropriate C (i. This known non-Turing-recognizable language can be any language for which non-Turing- recognizability has been proved in the textbook, in lectures, in class handouts, or in homework problems By definition, a language is decidable if there exists a Turing machine that accepts it, that is, halts on all inputs, and answers "Yes" on words in the language, "No" on words not in the language. L ∈ R iff L is decidable Commonly it means $\subseteq$, which is subset or equal. So D A I rejects the 15 May 2017 We discuss reductions which, in this context, are ways to transform instances of one problem P into instances of another problem Q in such a 12 Dec 2012 Proving additional languages are not decidable, by using reductions. 11. Proof: We use T to decide A. I. First of all, rename all the terminal symbols in the second grammar so that they don't conflict with those in the first. We will prove shortly that there are languages that are not semi-decidable If 〈M〉 ∈ L then there is no number k such that M accepts wk. By Rice’s theorem, we know that A is undecidable. We say that A is mapping reducible to B (denoted by A <= m B) whenever there is a computable function f: Sigma* -> Sigma* such that for all words w in Sigma*, w belongs to A if and only if f(w) belongs to B. Some languages are not Turing-recognizable. L 3 = fh M ij is a TM and j 3 g. There are two stacks i. Page 3. A theory is a set of formulas, often assumed to be closed under logical consequence. For example, a decision procedure for equality produces a decision of the equality of two things: either they are equal, in which case the procedure has constructed a proof that they are equal; or they are not, in which case the Sep 14, 2010 · The theme of this paper is computation in Winfree’s Abstract Tile Assembly Model (TAM). Oct 12, 2018 · The set of programs that run on a Turing machine and halt that are not decidable is a NULL set. Simulate M1 on w. Thus, an algorithm running in time O(n2) is ecient; an algorithm running in time O(2n) is not. Examples of decidable Languages Proposition 1 A DFA = {hB,wi : B is a DFAthat accepts w} is a decidable language. Problem 4: Enumerability and decidability Suppose that L is a language over alphabet {0,1}∗. For an infinite set S [[subset]. L is not decidable, then L' cannot be decidable. By the pigeonhole principle at least one nonterminal must be repeated on this path, since there are only k of them. 24. R inputs the pair , where M is a TM and w is a string. However, a direct proof can be done in a similar fashion as Question 3b, but with ATM and ATM;unary (instead of the complements). _ 3) one of L and L is acceptable, but not decidable; the other is not acceptable. , given one undecidable language (i. Draw the transition diagram for a DFA D 3 for not possible to pump 1’s and 2’s or 3’s and 4’s at the same time. (a) The complement Lis decidable. Prove that the following language is decidable: Formulate the following problem as a language and prove that it is decidable: Given a PDA and a string, determine if the PDA accepts the string. Sometimes ﬁnding an appropriate s is the hard part. Proof by diagonalisation that not all languages are Turing-recognisable. Of course, it is an open question whether P is properly contained in NP. The problem with adapting the proof for DFAs is that the class of context free languages is not closed under comp. • The classes of Turing-recognizable and Turing-decidable languages are different. Choose another language L1 over Σ∗ that is known not to be recursive. [2] PAL. to prove decidability of the ﬁrst order theory of reduction. It is easy to show that the set of even numbers is decidable by creating the relevant Proving a Language is Not Regular We’ve seen in class one method to prove that a language is not regular, by proving that it does not satisfy the pumping lemma. If a theory is complete, it is decidable (proof sketch: given a sentence \(A\), systematically generate the theorems of the theory; by completeness The language LC is not decidable. True or False: If 1 is a Turing recognizable language and . In particular, recognizable languages would coincide with decidable languages (which is verycoincide with decidable languages (which is very unrealistic!). Assume B is decidable. The language fhM;wi: Maccepts the input wgis not decidable. Decidable Problems for Regular Languages Theorem A DFA is decidable, where A DFA = fhB;wijB is a DFA and w 2L(B)g (Proof idea) IWe construct a Turing machine M to decide the problem. The theorem is stated in negative form, because this is how it is used, i. , for two predicates with the same set of attributes, p = q is true iff these symbols are interpreted by the same relation), that is, the strong equational theory of intensions, is not decidable, in general. The Bitcoin scripting language is a stack-based language similar to Forth. The notion of NP-complete is based on the following notion from computability theory: Deﬁnition 2. Let L₁ and L₂ be decidable languages over the same alphabet Σ. Proofs of Undecidability. Your language L is indeed undecidable. L() = (L(A) ∩ not(L( ))) U (not(L(A)) ∩ L()) Called Symmetric Difference If L(C) is empty then A & B gen same language Then use machine from Theorem 4. (computer science) describing a set for which there exists an algorithm that will determine whether any element is or is not within the set in a finite amount of time. Decidable and Semi-decidable Languages (Score: _____ out of 20 points) Let Sigma be an alphabet. is not) a member of the language. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the following diagram − Example 1. Prove that your DFA D 1 does indeed recognize L 1. , it takes only a finite amount of time to run on any input string). Therefore one way of showing that a language is decidable is by describing a Turing machine that accepts it. IEvery nite language is decidable: For example, by a TM that has all the strings in the language \hard-coded" into it IWe just saw some example algorithms all of which terminate in a nite number of steps, and output yes or no (accept or reject Undecidable definition is - not capable of being decided : not decidable. Regularity for this class was proved decidable in [3]. Indeed if L(M 1) = ;we have M 1 2=L0, while if L(M 1) = L(M 2) we have M 1 2L0. language, such that {} has property P. Contradiction is shown for all cases, proving that L is not a Regular Language. S is "respectable" in Hunter's sense (hence consistent and powerful), and "well-made" in the sense that its sets of wffs, axioms, and transformation rules are all decidable. In other words, we’re going to show that the following language LE is not decidable: LE = {"M": ε ∈ L(M)} We will show that if LE is decidable, so is H = {"M" "w" : w ∈ L(M)}. hangs) on a string that is (resp. I Function f is a reduction from A to B. Decidable and Undecidable Languages. One of Turing's great contributions is in formally defining Define the language A = {<B> | B is a TM in this model and <B> is not in L(B)} Suppose A were decided by a machine M in this model. • As a consequence of Theorem 4. To prove a language is undecidable , need to show there is no Turing Machine that can decide the language. Hence the property is non-trivial. Unfortunately, this doesn't work at all. Given set E of equational axioms So a problem is eciently solvable (a language is eciently decidable) if there exists an algorithm (such as a Turing machine algorithm) that solves it in time O(nd) for some constant d. Then for every language L2NTIME(t(n)) there is a constant csuch that L2TIME(2c t(n)). Show that if L1 were decidable, L2 would be as well. Proof To prove that LC is not decidable, we assume that it is decidable by the TMMC and show that this implies the existence of a TM M H that decides L H, which has been shown previously not to exist. If <M> ∈ L(M), then <M> ∈ A, proving that <M> is not in L(M). Show that the following languages are context-free. We prove several decidability and undecidability results for the satisﬁability and validity problems for languages that can express solutions to word equations with length constraints. We reserve the term algorithm for this class of problems. Here we show that there exists a language that is not even recognizable. ▷ The Church- Turing We'll prove this using Cantor's diagonalization method, a type of proof by 17 Oct 2019 algorithm works? – Note: The proof is not only to prove it works for a specific We have shown that a CFL is decidable and a CFG can be simulated by a TM. This claim is a special case of [14, Theorem 5. Therefore, by Rice’s Theorem, L0 is not decidable, a contradiction. Solution: This is in fact a simple test on applying the deﬁnitions and Theorem 7. The recursively enumerable languages (RE) are ones for which there is a TM that always halts accepting strings in the language, but for strings not in the language might halt rejecting the string or might run forever. 22. We know Xdoes not exist. The easiest is to use deﬁnition 3. Convert a tradtional k ~> Boolpredicate into a Predicate. Examples Decidable: , , , , 𝑃 , Decidability: Reduction Proofs (2) Steps of a reduction proof Given language L 2, where want to show that L 2 2=D(or SD) 1. In particular, solving a characterization problem requires a proof that a language satisfy-ing some decidable property belongs to the class, which usually yields a canonical construction for the language. Y 3. By Church's thesis, it doesn't matter which machine model we assume, or what language we use to write the program. (Finding an algorithm In [8], a decidable quanti er-free fragment of an array theory that allows a restricteduseofapermutation predicateisstudied. Then, we can prove the theorem. a. Termination is to reach the normal form in finite time. What does decidable mean? Information and translations of decidable in the most comprehensive dictionary definitions resource on the web. This makes sense even if B is not decidable! (We do not assume that the oracle B is a computable set!) Decidable languages Semi-decidable languages Co-semi- Careful: To prove P2 undecidable, we must not reduce P2 to some known undecidable prob-lem P1 (by converting instances of P2 into in-stances of P1 that have the same answer), as we would then prove the vacuously true and thus useless statement “if P1 is decidable, then P2 is decidable”. •Determine and prove whether sets are countable. To choose If the halting problem was decidable then every recursively enumerable language would be recursive Theorem: The halting problem is undecidable Proof: Assume for contradiction that the halting problem is decidable There exists Turing Machine that solves the halting problem H H M w YES Mhalts on w Mdoesn’t halt on NO w We can prove this using a counting argument. This shows that just because a Turing machine’s language is decidable, it’s not necessarily the case that the Turing machine itself must be a decider. The structure is also definable in Presburger arithmetic. To cover various examples in programming language theory, we combine and extend both syntactical and semantical results of second-order computation in a non-trivial manner. reject. False. 4. is decidable and let TM )be its decider. If any of these numbers produce a remainder zero, then it goes to the “Rejected state”, otherwise it goes to the “Accepted state”. The de nition is similar to that of a Deterministic Finite Automaton, or Push Down Automaton, but it can do much much more. 8. Construct C, the product automaton of A and B. Are there problems that cannot be solved by any algorithm? By definition, a language is decidable if there exists a Turing machine that and answers "Yes" on words in the language, "No" on words not in the language. Note that trying to use "complementation" to solve (a) will not work, because the complement of a context-free language is not necessarily context free. EQ. Contradiction. Define decidable. The atomic formulas over this language are equality over string terms (word equations), linear inequality over the The language Preﬁx(A) might not be decidable, even if A is de-cidable. If x = ", then run M on input w and accept if M accepts w. On the other hand if the languages are not equal, the machine Mwill sooner or later ﬁnd a string which is in one language but not in the other one and accept. A language A is mapping reducible to language B, written A ≤≤≤≤mB, if there is a (10 points) Let A and B be two languages. Recursive and Recursively Enumerable Languages, Unrestricted Grammars. " 4. is decidable. The always-false predicate. Since each language is RE, for each language we can construct a Turing machine with a single tape that accepts all inputs in that language. Instead of the heavily case-based approach of Banks and Cilleruelo-Luca-Baxter, we used a decision method: we recoded the problem as a formal language theory problem, and then used the fact that this formalism has a decidable logical theory associated with it. COROLLARY: Some languages are not Turing-recognizable. IScan the input string repeatedly. Run M 2 on w. In this case you are done with your observation that every $\mathsf{DSPACE}(n)$ language is decidable. @#@$$$p :: Predicatek) Source#. 5. I'll present an example of a decidable language, followed by a general result about decidable languages. 2 Nov 11, 2016 · Decidable equality. A description of a TM M which decides A DFA. In particular, we prove that it is undecidable to determine whether a given Turing machine accepts a given input string. 4. Then there is a decider Turing machine D 2 that decides Halt 2. $\endgroup$– PrudiiArca23 hours ago Exercise 7. 𝑇𝑇𝑇𝑇𝐴𝐴. 1: If L 1 is not recursive and there is a reduction from L 1 to L 2 then L 2 is not recursive. 7 Draw the transition diagram for a DFA D 1 for the language: L 1 = { w ∈ {0,1} * | w does not contain the substring 11 }. Finally we prove in section 8 that typing is decidable. ) (d) Prove by diagonalization that there exists a recursive set that is not accepted by any LBA. This is similar to what we have seen before, except that now we are working with Turing machines. There exists some language which is in NP but is not Turing decidable 3. Suppose we have two context-free languages, represented by grammars with start symbols and respectively. There are three methods you may use to prove this is true. In Idris, decision procedures are really useful. Nov 29, 2006 · Ah, there we go. So, it is natural to expect that A TM is NOT decidable. The recursive languages are closed under union, so A∪ Bc is recursive. The Pumping Lemma for regular languages. Find out whether the following problem is decidable or not −. To prove that a given language is non-Turing-recognizable: Either do both of these: Prove that its complement is Turing-recognizable. , and L0 ≤ m L for every c. In other words, a language is decidable exactly when both it and its Nov 01, 2008 · Corollary 11 There exists a polynomial time decidable set F [[subset]. The book is based on what psychologists call a "model. Proving additional languages are not decidable, by using reductions. If s = w, accept 3. A TM M which decides L works as follows: M="On input w 1. Suppose a language L is enumerated in lexicographic order by an enumer-ator E. 03/03/2010 20 Guide To Classifying Languages Claim: L is decidable. Returning to our example, suppose that w is a string so that |w| > 24 = 16 and suppose also that A is the nonterminal repeated in the parse tree for w. The problem has one input, the regular expression R, and should therefore be expressed as the language RU 110 def= fhRijRis a reg. The property P is about the language recognized by Turing machinesif whenever L(M)=L(N) then P contains (the encoding of) M iff itcontains (the encoding of) N. L is said to beTuring-decidable(Recursiveor simply decidable) if there exists a TM M which decides L. 25 Consider the language J = fwjeither w = 0x for some x 2A TM; or w = 1y for some y 2 A TMgde ned in problem 5. P is decidable. If M1 accepts, then ACCEPT w. $\endgroup$ – Emil Jeřábek Apr 24 at 9:55 1 $\begingroup$ @Emil the input is still a context free grammar but we ask if there exists a context free grammar that is unambiguous and language Nov 11, 2014 · You need to show that, given a Context Free Grammar G, one can check whether the language L(G) recognized by G is included in a*b*. This argu-ment, however, does not give us an explicit construction. bar] N, define bounded automata is decidable. We call ya proof (or certi cate) of xbeing in L. rejects . This question hasn't been answered yet language). On input (M,w), make a new TM N that on input w marks the leftmost tape cell and then simulates M(w) (as tho the leftmost cell was not there). To illustrate the problem of typing consider the following situation. E. Theorem 5. The base case is n = 2 is by hypothesis. It provides primitives to support modeling in decidable logics, and also modular refinement techniques the makes it possible to separate the verification effort into local, decidable problems. Given a DTM M deciding a language L = L 3 Apr 2006 is decidable. So, here the answer could be made by ‘Yes’ or ‘No’. As a consequence, the HOM problem is proved decidable for the partic- A formal language is said to be partly decidable (or recursively enumerable, or Turing-recognizable) if there is an algortithm (again, epitomized as a Turing machine) that always halts with YES if 1. , problem) to show that the latter is also undecidable. Therefore, you may provide a decider Turing machine M such that L(M) = L to prove L is decidable. Otherwise, accept. decide against (someone or something) 1. Proof by diagonalisation that the acceptance problem for TMs (ATM) is not decidable (but is Turing-recognisable). Moreover since decidable languages are closed under complement, L is also Turing 1. To show that L1 is undecidable: Identify L2 that is undecidable. I read in wikipedia that RE languages are closed, but I didn't find any proof and also I didn't find anything about R languages. Build a TM *that decides !!": *= on input 4 1. In decidable ⊆ RE ⊆ all languages our goal: prove these containments proper regular languages context free languages all languages decidable RE 7 January 29, 2020 CS21 Lecture 10 Countable and Uncountable Sets •the natural numbers N = {1,2,3,…} are countable •Definition: a set S is countableif it is finite, or it is infinite and there is decidable (comparative more decidable, superlative most decidable) capable of being decided. In fact, it's not even Turing-recognizable! Closure properties of Turing-recognizable and Turing-decidable languages. (Sipser, Problem 3. , a C such that LC = L), prove that (1) C µ RE, (2) C 6= RE, and (3) C 6= ;. If we consider the two formulae above, and we Undecidable Languages there is no Turing Machine which accepts the language and makes a decision (halts) for every input string undecidable language = not decidable language There is no decider: (machine may make decision for some input strings) Prove the following theorem. semi-decidable languages that are not decidable. – But D does not exist! • Conclusion. If L is a language in NP, L is Turing decidable Which of the above statements is/are True? If not, reject. Theorem (Turing, 1936). strings of 0's and 1's such that no prefix has two more 0's than 1's, not. So, assume that Halt 2 is decidable. Proof The definitions indicate that not all languages are decidable. Theorem 1. The Halting Problem is not Turing-decidable. Run M on w. 181-182: Language L is decidable ⇐⇒ both L and Lc are recognizable (Lc is. wedge construction, and gives a new characterization of decidable languages of natural numbers in terms of self-assembly. Prove that ECFG is a decidable language. I have read much of Chaitin and never seen a hint that he has considered these matters. The automation features have been de- We prove undecidability of the above language Halt 2 by showing that if it were decidable then so would the first version of the halting problem, Halt, which we previously proved to be undecidable. The RE languages are not closed under complementation. So how do we prove this? First we'll prove this using the TM/language framework. , To prove a language is undecidable, need to show there is no Turing Machine that can decide the language. e, this idea provides a recognizer but not a decider. Department of Software Conversely, because 01 is decidable, if A m 01, A is also decidable by Theorem 5. (a)On input w: 1. If )rejects, reject. IScan the input string hB;wi, determining whether the input constitutes a valid DFA and string w. A decision problem P is decidable if the language L of all yes instances to P is decidable. (a) Closed under union. 1. A language L is decidable if and only if both L and L are Turing recognizable. -complete iﬀ L is c. We consider two cases, the ﬁrst in which Bis in not Cand the second in which it is in C. • But the other direction does not hold---there are languages that are Turing-recognizable but not Turing-decidable. (10 points) True/False. L 5 = is decidable Proof: Construct a new DFA C from A and B that Accepts only those strings that are accepted by either A or B, but not both i. is a decidable language, then 5 < must be Turing-recognizable. 22 we get a hold of languages that are not Turing-recognizable ( #is the complement of A): Theorem C Let A Y -* be a Turing-recognizable language that is not Turing-decidable. Show that non-regular languages are NOT closed homomorphism. Then there exists a TM M B that decides B . Corollary: If L is not decidable, then either L or L is not enumerable. Does not hold if 5 0 (and . and how to prove countability and uncountability • Understand how diagonalization is used to prove the existence of explicit undecidable languages (𝐴𝐴. To reach a conclusion or form a judgment or opinion about by reasoning or consideration: decide what to (6) From the options given below the statement, which is not necessarily true if X1 is the recursive language and X2 and X3 are the languages that is recursively enumerable but not recursive is (A) X2 ∩ X1 is recursively enumerable (B) X2 ∪ X1 is recursively enumerable (C) X2 – X1 is recursively enumerable (D) X1 – X3 is recursively We prove several decidability and undecidability results for the satisfiability/validity problem of formulas over a language of finite-length strings and integers (interpreted as lengths of strings). What you're proposing to do is to simulate a Turing machine on some input, and ask whether it accepts/rejects that input in To prove that a given language is decidable: • Construct an algorithm that either reject or loop on any string not in the language. Can a TM just simulate P on w, accept if it accepts and reject o. Show that the set {x|∃y|y|= |x|2 Careful: To prove P2 undecidable, we must not reduce P2 to some known undecidable prob-lem P1 (by converting instances of P2 into in-stances of P1 that have the same answer), as we would then prove the vacuously true and thus useless statement “if P1 is decidable, then P2 is decidable”. Please clearly show all three steps. To construct an example that illustrates that this is so, let us ﬁrst take B f0,1g to be any language that is Turing recognizable but not decidable (such as HALT, assuming we deﬁne this language with respect to a binary string encod-ing of DTMs). m](A) is a finite union of finite intersections of EqualToais a predicate that the input is equal to a. 7. That is, all words in the language are accepted by the TM. Slide 58 Chomsky Language Class Hierarchy Today we will explore a mechanism by which we can prove that a language is not decidable. not necessarily decidable. The always-true property, which contains every RE language. I would really apreciate any help. net dictionary. – We assumed A. One can also prove it directly by showing that each morphic preimage and also each derivative of a language from the class [U. Then, this follows from the closure of Turing-recognizable languages under concatenation (this was again a Decidable, Undeciable, and Beyond Exposition by William Gasarch 1 De ning Decidable How to pin down what is meant by \computable?" This de nition is moti-vated by actual computers and resembles a machine. Finally, I conclude that LC is not recursive. Page 5. Since $M$ is a dfa, we already have the Turing Machine and just need to show that the dfa halts on every input. In fact, Theorem 1 better explains the situation. L e t w be a binary string, and Mi be a TM. (b) Prove that for every language L, there is a decider M that accepts every string in L and Dec 07, 2012 · Yes, any language in P or NP is decidable. A partially completed DFA that accepts this language is shown below. In this section we make precise the concepts used in this paper and derive prelimi- nary results. First recall that solving a problem can be viewed as recognizing a language (see Problem Solving as Language Recognition). 18 Oct 2018 Are any languages undecidable, and if so, how do we prove it? The Turing machine U actually recognizes HALT, but it does not decide it. Notes: Choosing a s that can be pumped proves nothing. Bottom line: For every \strictly semi-decidable language Decidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. decidable by a TM). strings of 0's and 1's whose no of 0's is divisible by 5 and no of 1's is. Solution: 1. So take B to be a language which is RE, yet Bc is not RE, (for example, let B be the language corresponding to the halting problem). Prove that a language L is recursively enumerable if and only if it can be expressed as L = fxj there exists y such that hx;yi 2 R g where R is a decidable language. By the de nition of J, w 2A TM i f(w) 2J. (a)Any subset of a decidable set is decidable. Then we used publicly-available software to prove our result. •Use diagonalization in a proof of undecidability. over f0;1gand there is w2L(R) such that w= 11u0 for some u2f0;1gg 2. We first review a simple, well-known tile assembly system (the “wedge construction”) that is capable of universal computation. In other words, you must prove that every language of this form is RE, and that every RE language has a related decidable language R that allows it to be expressed exactly as De nition: A theory is decidable i f#AjA2 gis recursive. Then we will prove something stronger: There are semi-decidable (r. (Hint: You need to be able to detect after a ﬁnite time if the machine is in an inﬁnite loop. (20 points) Show that the set of decidable languages is closed under intersection. Technically speaking, our characterization is (exactly) the ﬁrst main theorem from Lathrop et. We know we can type an application if we can nd an arrow type for fwhose domain types a. 2 is decidable, then L 1 is decidable. is 6 798. Dec 12, 2012 · Proving additional languages are not decidable, by using reductions. Let Abe any regular set. ) * (3) All the King’s Horses, and All the King’s Men… Let ALLDFA = {<A> | A is a DFA and L(A) = Σ*} Describe in English what the language ALLDFA consists of. Now LG( )=Σ *if and only if M does not halt when started on blank tape. To make this method correct and decidable, we must ensure that the rewriting rules are terminating and confluent. First, we copy the input to every tape. 7 The Turing machine that decides (solves) a problem answers YES or NO for each instance of the problem. M is a The proof is left as an exercise. This machine is guaranteed to halt since Mis a decider, and accepts exactly the strings in L . not in the Language. Lemma. You can do this by writing a context free grammar or a PDA, or you can use the closure theorems for context-free languages. (a)Show that the class of decidable languages is closed under union. Proof: In order to derive this from Theorem 9 it is sufficient to observe that there exist shifts whose language is co-c. • This algorithm may “call” any To prove that a given language is undecidable: • Construct a (mapping) reduction from another language already known to be undecidable to the given language is not Turing-decidable. is decidable, however A TM is not decidable. ➭ I. We 4 Oct 2016 On the other hand, the class of semi-decidable languages is not closed under complementation. 18) Show that a language is decidable iﬀ some enumerator enumerates the language in lexicographic order. I have managed to prove thet CFL are't closed under division. 3. , problem) we reduce it to another language (i. is a DFA that . 2 of the textbook extends this to a theorem giving that a language Ais decidable if and only if it’s both Turing-recognizable and co-Turing-recognizable. Proof by Reduction 0. Hence the result they actually prove is that it is undecidable whether a given context-free grammar generates an inherently ambiguous language. Intuitively, for recognizability we allow our TM to run forever on inputs that are not in the language, while for decidability we require that the TM halt on every input. The language A. F]) is not decidable. Consider. _ 2) neither L nor L is acceptable (r. , we can easily decide whether a regular language is empty, finite, all of $\Sigma^*$. for input hB,wi run B on w; if B accepts w, M accepts hB,wi; else M rejects hB,wi. Prove that RU 110 is decidable. Make the final states of C be the pairs where A-state is final but B-state is not. Give an example of a DFA Astallion that in in ALLDFA. Lecture 17: to show there is no Turing Machine that Since X does not exist, but Y could be used to make X Otherwise (N, y) does not halt, then M will not halt for any input string no matter how long it is, thus M is not in L. BoolPred:: (k ~> Bool) -> Predicate k. This method works often but not always. Follows later. (e) If A is recursive and B is Recursively enumerable, then A∪ Bc is: None of the above Proof. It must either reject or loop on any string not in the language. Show how to use Y to make X. Note: you must prove set containment in both direction, that L(D 1) ⊆ L 1 and that L 1 ⊆ L(D 1). Prime numbers = { Proof by contradiction. Or: Construct a (mapping) reduction from another language already known to be non-Turing- recognizable to the given language. Prove or disprove: Lmust be Turing decidable. L e t L e denotes an empty language, and L ne denotes non empty language. If L is decidable then it is Turing recognizable. We will show that RA is undecidable, and use this to prove that in fact every sound theory (over the language L Nov 26, 2009 · Write Regular Expression for the language that have the set of all. Choose language L 1 such that L 1 L 2 L 1 2=D(or SD) 2. If a language is not even partially decidable , then there exists no Turing machine for that language. For example, 001110 and 011001 are in the language, but 100010 is not. DFA = { 〈 D,w 〉 | D . Some Languages are Not Turing-recognizable. Proof of the theorem: If L is Turing-recognizable and its complement is, then L is Turing decidable. 4 1. Step 1: Show how your machine works on an input string Step 2: Show that the language of your machine is IDFA Step 3: Show that your machine is a decider (i. Imagine trying to type fain di erent contexts. The first proof is non- constructive, using Cantor's diagonalization. How to use undecidable in a sentence. Important detail: M and M must be run in parallel, not sequentially! See next slide. Digression: How to Run Two Nevertheless, no general proof exists. 4 Choose a In particular, M does not accept the string ILLINI. For a direct proof, Sipser gives a reduction from the language A_TM. Undecidable Problems – The problems for which we can’t construct an algorithm that can answer the problem correctly in finite time are termed as Undecidable Problems. Jul 17, 2012 · Abstract: Condon and Lipton (FOCS 1989) showed that the class of languages having a space-bounded interactive proof system (IPS) is a proper subset of decidable languages, where the verifier is a probabilistic Turing machine. expr. And this is not just a remote historical issue but is totally current. Pumping Lemma for Context- Free Languages. 7 on page 172). THEOREM 4. The construction of TM M A using TM M B establishes that A is reducible to B . Computability and Complexity. Turing Machines that halt in all inputs are called deciders. The complement of every Turning decidable language is Turning decidable 2. Confluence ensures the existence of unique normal forms. He constructs a decider R for A_TM out of a decider for REGULAR_TM, as follows. A language is co-Turing-recognizable if it is the complement of a Turing-recognizable language The complement of a language is the language consisting of all strings that are not in the language. L1 should be chosen to be similar to L2. De nition 2. 10. Let Lbe a decidable language and Mbe the Turing machine that decides L. > Such languages are not Turing recognizable. The extensional equality theory of predicates and functions under higher-order semantics (e. 2 Another pr oof that A T M is not decidable Recall A T M = {< M ,w > |w decidable T-recognizable Complement of T-recognizable = co-T-recognizable TM TM Check-in 8. Formulate the following problem as a language and prove that it is decidable: Given a DFA, determine if it accepts some palindrome. therefore in section 4 we show that the latter are decidable. Thus, MC cannot exist. Corollary The complement of HALT is not CE. So, deciding whether or not &4=&$ # with TM is recognizable but NOT decidable Can often prove a language L is undecidable by proving: if L is decidable, then so is A TM We reduce A TM to the language L A A Non-enumerable Language Earlier we saw Theorem: If L and L are both enumerable, then L is decidable. That is, L(D 1) = L 1. Finding a decidable language notin $\mathsf{DSPACE}(n)$ is only necessary, if $\subset$ is meant as $\subsetneq$, ie. De nition: If and 0are theories, then is an extension of if 0. The relation R is Turing-decidable if and only if R is a recursive relation. In this paper, we show that if we use architecturally restricted verifiers instead of restricting the working memory, i Logic and Language. For example, you could show that L is the union of two simpler context-free languages. 32-23. Set up a proof that claims that L L is regular, and show that a contradiction of the pumping lemma’s constraints occurs in at least one of the three constraints listed above. 1. Decidable Problems Concerning Context-Free Languages ◇As always, languages must be defined by instances of another. Exercises. The L EMPTY language (consisting of those encodings of TM's M such that L(M) = ∅) is not Turing-decidable. The Question: Are there languages that are not decidable by any Turing machine (TM)?. Nov 09, 2017 · (decidable), or prove from scratch that the problem is hard (undecidable) (e. Although it might take a staggeringly long time, M will eventually accept or reject w. Theorem. THE PROBLEMS: 1. To determine which language contains each input, we construct a Turing machine with k tapes. is, in particular, recognizable. However, suppose Dwhere decidable and let T be a Turing machine that solves D, then T(hTi) halts and accepts if and only if T(hTi) does not halt and accept, which is a contradiction. Then, briefly justify why your construction is correct. 2 Reductions Note that the key to this proof is to use the fact that A TM is not Turing-decidable. 22 A language is decidable iff it is Turing-recognizable and co-Turing-recognizable. Prove the following properties. Counterexample. First, 2 definitions: * NP is the class of languages that can be recognized by a nondeterministic Turing machine (TM) in For an undecidable language, there is no Turing Machine which accepts the language and makes a decision for every input string w (TM can make decision for some input string though). 1HALTTM= { ¢M, w²|M is a TM and halts on input w } is Turing-recognizable, but not decidable. To show that a language is decidable, we need to create a Turing machine which will halt on any input string from the language's alphabet. Run Mon hwi 2. A property about Turing machines can be represented as the languageof all Turing machines, encoded as strings, that satisfy that property. 3: is;, which is not recognizable. The language L is But since assertions are expressed in predicate logic and predicate logic is not decidable for arbitrary expressions the theorem prover in the Albatross compiler cannot prove all valid assertions. Let L NP be some recursively enumerable language that does not have the property P, and let M NP be a Turing Machine such that L[M NP] = L NP We will create a machine M′ such that M′ has property P if and only if Mdoes not halt on w. We prove that RU 110 is decidable by constructing a decider for it. To prove that a language is not regular, use proof by contradiction and the pumping lemma. • In order to prove this corollary, recognized by a given DFA A is empty or not. (a) L = a ncb (a) Prove that there exists a Turing machine M whose language L is decidable, but M is not a decider. ) languages that are NOT decidable We'll prove Gödel's theorem for a system S of formal arithmetic. Problem 7. –L s iTuring-decidable if there is some TM that decides L. (and no, it’s not another Pumping Theorem…) The primary technique we will explore is reduction. Solution Oct 17, 2014 · Church-Turing thesis; examples of decidable languages. Theorem: Let t(n) nbe a function from natural numbers to positive reals. decidable. We construct a TM Mthat L= L(M) as follows 1. Find out whether the following problem is decidable Problem 3. It implies that A TM m J 1. reductions from languages unknown to be decidable (Section 16. True. Given all this, we give the following decider D PAL for PAL: D PAL Is the Halting Problem Decidable? Can we build a program that solves the halting problem for any program? It is important that we pose the problem with respect to any program, not the handful of programs we know. (more languages than machines) We say that a language is co-Turing-recognizable if it is the complement of a Turing-recognizable language. By deﬁnition it follows that EQ The language you use to explore a computational problem can impact the clarity or readability of your code but not whether a solution to a problem exists. If L and M are regular languages, then so is L – M = strings in L but not M. 1 (Decidable Languages) Let Land L0be decidable languages. Assume fhM;wi: Maccepts the input wgis decidable, then the language D = Q: What would happen if we prove that A TM is decidable? If A TM were decidable, the language of any TM would be decidable. is not recursive. Use Reduction To Prove That The Language Is Not Decidable ( And Are Turing Machines). If R accepts, accept; if R rejects, reject. If is not decidable, then or ̅ is not Turing-recognizable. Then Bn = Bn 1B BB B (where the rst inclusion is by the induction hypothesis). Section 4. A language is Turing-decidable if it halits in an accepting state for every input in the language, and halts in a rejecting state for every other input. v. It is clearly a property of the language of M. [Note. It is easy to do something similar with time-bounded computations. This is hard: requires reasoning about all possible TMs. language L0. typeBoolPred(p :: k ~>Bool) = (EqualToTrue. If these sets are decidable, then the set of proofs is also decidable. Are there problems that cannot be solved by any algorithm? Consider the language: ATM = {<M,w> | M is a TM and M accepts w} NOTE: <A,B,… > is just a string encoding the objects A, B, … First of all, the undecidability of REGULAR_TM follows immediately from Rice's theorem. 22 on pp. A language is Turing-recognizable (or recursively enumerable) if it is recognized by a TM. To surmount this fundamental diﬃculty, the key idea of our argument is to prove that for any ﬁxed ﬁnite collection A of pieces, the reduct substructure ChA of positions using only at most the pieces of A is not only computable, but in the restricted language of chess is an automatic Jun 18, 2012 · The proof proceeds by showing that the mate-in-n problem is expressible in what we call the first-order structure of chess \(\mathord{\frak{Ch}}\), which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable. Prove the following: 1. EDFA = 1<A> : A is a DFA and L(A) = ∅l. L = { <M> | M is a DFA that accepts infinitely many strings } In other words, the computational problem of determining whether a given DFA accepts an infinite language or not is decidable. TM. We demonstrate how to prove decidability of various algebraic theories in the literature. The function f is Turing-computable if and only if f is a recursive function. TM was decidable. 6 (page 142). UNDECIDABLE. Therefore the programmer has to provide the proof steps which cannot be done by the compiler automatically. A recognizer of a language is a machine that recognizes that language; A decider of a language is a machine that decides that language; Both types of machine halt in the Accept state on strings that are in the language ; A Decider also halts if the string is not in the language ; A Recogizer MAY or MAY NOT halt on strings that are not in the Undecidable Languages The Question: Are there languages that are not decidable by any Turing machine (TM)? i. Express the problem as a language RU 110. Accepting, Deciding and Languages Recursive Languages are decidable. Proposition 2 A NFA = {hB,wi : B is an NFAthat accepts w} is a decidable language. , have a semi-algorithm) if the language L of all yes instances to P is RE. A language L is c. Even with advances in quantum computing, we will never be able to create a general-purpose debugging program. of RE languages). 2) This exercise concerns TM M 1 whose description and state diagram appear in Example 3. • Theorem 2: If L is Turing-decidable then L is Turing-recognizable. WaitforEtoprintastrings. Hence Maccepts if and only if L(G 1) 6= L(G 2) and it is so a recognizer for EQ CFG. Proof: in class Theorem. Run R on input hM 2i. Prove that L₁ ∪ L₂ is also decidable. If not, then reject. an input not in L. Key Concepts Turing machines, recognizable languages, decidable languages, undecidability 1. Undecidable languages are not recursive languages generalized to prove Rice’s Theorem: • Let P be any non-trivial property of Turing-recognizable languages ‣Non-trivial means P is true of some but not all • Then { M ⃒P is true of L(M)} is undecidable • Examples of undecidable properties of L(M): ‣L(M) is empty, non-empty, finite, regular, CF, mB, if there is a computable function f : ! such that w 2A if and only if f (w) 2B. Run )on 4,$ #. A language L is decidable if and only if L is decidable. That is, show that if L1 and L2 are decidable languages, then L1 intersection L2 is a decidable language. undecidable languages •We first introduce the diagonalization method, which is a powerful tool to show a language is undecidable •Afterwards, we give examples of undecidable languages that are –Turing recognizable but not decidable –Non-Turing recognizable Objectives Definition of decidable in the Definitions. Partially decidable problems and any other problems that are not decidable are called undecidable. An algorithm is defined by the existence of a TM that implements the algorithm. A language is Lsemi-decidable if there exists a Turing machine V L (where V stands for "veri er") halting on every input such that for all x2 , x2Lif and only if there exists a nite string y(over a possibly di erent nite alphabet) such that hx;yi2L(V L). al. 27 Oct 2011 Proof. The recursive languages are closed under complementation, so Bc is recursive. We did not need to examine a supposed Turing machine that recognizes A TM and 2. Since K and L are decidable languages, it follows that there exist turing machines M K and M L that decide the languages K and L respectively. Then #is not Turing-recognizable. Examples of how to use “decidable” in a sentence from the Cambridge Dictionary Labs Decidable Languages Costas Busch - LSU LSU * We will prove that there is a language : is not Turing-acceptable (not accepted by any Turing Machine) is Turing of semi-decidable languages as follows. ? R. " It's a way of categorizing things that How can you prove a language is decidable? 3. The language ATM is undecidable. 1] or alternatively of [9, Theorem 1]. If it accepts, accept Proof. Hence Mwill only loop if L(G 1) = L(G 2). 11 on page 260. Exercise 16. Please prove that the following language is decidable by designing a decider. 4), and (iii) { 〈A〉 | A is a DFA that does not accept any string with odd 1s}. Apr 10, 2018 · 1. In case the string does not belong to the language, the algorithm either rejects Closure Properties of Decidable Languages Decidable languages are closed under ∪, °, *, ∩, and complement Example: Closure under ∪ Need to show that union of 2 decidable L’s is also decidable Let M1 be a decider for L1 and M2 a decider for L2 A decider M for L1 ∪L2: On input w: 1. All strings of length less than 3 are also in the language. p = n ∪ ERR, where ERR is the easy to decide language: HALTTM is not decidable: counter-assumption: HALTTM = L(MHALT). On words not belonging to the language, the computation of the TM either rejects or goes on forever. For each of the following languages, state whether each language is (I) recursive, (II if L is decidable, then so is ATM We reduceATM to the language L ATM ≤ L We showed:ATM ≤ HALT TM Mapping Reductions f : Σ* →→→Σ* is a computable functionif there is a Turing machine M that halts with just f(w) written on its tape, for every input w. every other language does not contain it. 13 / 16. We will prove later that there are decidable languages that are not in NP; in fact, there are decidable languages than cannot be decided by any Turing Machine that runs in exponential time. If the simulation halted within that number of steps, then M halts. decidable if the Halting problem is decidable. This fact provides an important tool for showing undecidability; for example, if HP ≤ m L, then L is undecidable. It proves that Turing-recognizable languages are closed under union. tr. EDFA is a decidable language. We then extend the wedge construction to prove the following result: if a set of natural numbers is decidable, then it and its complement’s canonical two-dimensional Most decision questions about regular languages turn out to be decidable, e. If L is nite, then of course it’s decidable, so we suppose that L is in nite. If L(M j ) = Ф then Mi does not accept input then w is in L e . Proof: Let A and B be DFA’s whose languages are L and M, respectively. The busy-beaver function is not Turing-computable. 4/11. Nov 20, 2019 · By checking the production rules of the CFL we can easily state whether the language generates any strings or not. , but not decidable. Apr 01, 2019 · The Five Love Languages is a perennial seller, and has made its way around the internet as a quiz. Express the problem as a language L2 over some alphabet Σ∗. " 2. M decides a language when it halts (resp. We show that J is undecidable and J m J. If we can use M B as a sub-routine to construct a TM M A that decides A ,we ha ve a contradiction. DFA. At any rate, the bounds on the normal forms of well-typed terms are so astronomical, that the decidability theorem is mostly academic at this point anyways. 3 Prove that a language Lis decidable if and only if some enumerator enumerates L in lexicographic order. A is decidable A is also Turing-recognizable W e use contradiction. Prove that its complement is undecidable. I need to prove or disprove that R languages are closed under divison. Let R be its recognizer, clearly its running time is in nite =) hRi62A =) A 6= fhMig. Oct 17, 2014 · Proving additional languages are not decidable, by using reductions. If <M> is not in L(M), then <M> is not in A, proving that <M> ∈ L(M). I The idea here is that if B is decidable, then A must be decidable, too. If &4=∅, $ #and 4will have the same language (since &$ # =∅). Assume, for a contradiction, that TM T decides the language. A useless state in a Turing machine is one that is never entered on any input string. Proof: – The set Example 1. (Hint: Look at the proof for EDFA to get an idea. Then choose τ to express the similarity between the two languages. 2 Problem I. Theorem 15. Answer: One direction is easy. Prove that ALLDFA is decidable. Meaning of decidable. So the conditon Let's prove L recognizable again by finding an enumerator. decidable synonyms, decidable pronunciation, decidable translation, English dictionary definition of decidable. (b) Closed under intersection. 3 From what we’ve learned, which closure properties can we prove for the class of T-recognizable languages? Choose all that apply. (b)Any subset of a recognizable set is recognizable. Proof by Reduction. Context A problem is decidable if some Turing machine decides (solves) the Either: _ 1) L and L are decidable (recursive). Suppose that Lis sorted. , Y = a TM that can decide B) Y 2. Rice’s Theorem: For every other property P, L P is undecidable. •Use counting arguments to prove the existence of unrecognizable (undecidable) languages. In order to prove that KL is decidable, we can construct a turing ma- By the way, as it is implemented, Coq does not have decidable type checking, as it normalizes the body of fix statements before attempting to type check them. undecidable theory while propositional logic is a decidable theory. Every decidable language is Turing-Acceptable. PDA = { 〈 P,w IVy’s language is designed to make it practical to reduce all proof obligations to statements in decidable logics. 9. De ne the reduction Rfrom L Nov 20, 2019 · An undecidable language maybe a partially decidable language or something else but not decidable. If &4≠∅, $ #and 4will not have the same language. 2 Rice's theorem:Any nontrivial property about the languagerecognized by a Turing machine is undecidable. We will show there is no onto function from the set of all Turing Machines to the set of all languages over {0,1}. Then given a string a Turing machine that accept the language starts the computation. 2 Note that step 2 is guaranteed to terminate, because Mdecides (not just recognizes) L. A second method (which also doesn’t always work), is by using closure properties of (1)Let B be any language over the alphabet . We prove by induction that Bn B for n 2. (Works for any Σ) Hence there are languages that have no decider. Construct TM $ #on input /: 1. This means that there exists an algorithm that halts eventually when the answer is yes but may run for ever if the answer is no. Lecture 17: Proving Undecidability 6. If x 6= ", accept. 6. Page 23. On input x, M simulates ( N, y) for length ( x) steps. If )accepts, accept. To do so, suppose that you have methods inL1 and inL2 matching the above conditions, then show how to write a method inL1uL2 with the appropriate properties. This means that for the latter there are decision pro-cedures which for any formula decide whether it is valid or not — and CP° 1 in fact is such a decision procedure — while for the former such decision procedures do not exist in princi-ple. Solution: Lneed not be decidable, as is implied by Question 3b below. from lecture) • Know that a language is decidable iff it is recognizable and co-recognizable, and understand the proof Obtaining a decidable characterization for a class is considered as a way to get a ﬁne understanding of the class. also languages A ýB and A @B are Turing-recognizable. 1 Languages that Are and Are Not Context-Free 1. , given a representation of a regular language (either as a regular expression or as a finite state automata of some flavor) are decidable. T is able to prove all true sentences φ expressible in the language of the theory, due to the fact that either φ is true in T or ¬φ is true in T, then T is decidable. If 𝐽 is undecidable and 𝐽≤𝑚𝐽, then both ̅ 𝐽 and 𝐽 ̅are not Turing-recognizable. 1The proof of this is similar to the analysis in Question 1b. – They are not decidable and even they are not Turing recognizable. Prove that the following language is undecidable: A = fhMijL(M) 2TIME(n)g: Solution: L 1 = f1g2TIME(n) =)A 6= ; A TM is not decidable. If it accepts, accept. Yes, Holly was a strong candidate, but we ultimately decided against her for the job. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). Note: Remember that CFLs are not closed under intersection and that E TM is not decidable! 4 Nov 11, 2013 · Informally, being decidable means that there is a mechanical procedure which enables one to decide whether an arbitrary given sentence (of the language of the theory) is a theorem or not. ). intersection Solution: Proof. Jan 18, 2018 · Here it is known that the intersection of two recursive languages is a recursive language, then can't we say that its decidable that intersection will be recursive one? commented Jan 18, 2018 nikhil_cs -Turing recognizable languages are not closed under complement. What is lexicographic Question text at time of writing: > How do you prove that a Turing-recognizable language is not closed under complementation? Your question is not terribly well worded: it is technically nonsense, but could intend at least two or three things. Otherwise, it is not derivable by the equational logic. The following theorem does just that. 2. Proof Idea Let M be the TM that does the following: “On input <B,w>, where B is a DFA and w is a string: 1. L 1 = fh M ij is a TM and there exists an input on which halts in less than jh steps g. Let L 1 and L 2 be decidable languages and M 1 and M How do we know decidable? •Simplify problem at each step toward goal •Can prove formally –number of remaining symbols at each step Showing language is Turing recognizable but not decidable is harder 15 Many equivalent variants of TM •TM that can “stay put” on tape for a given transition •TM with multiple tapes A problem is called partially decidable, semi-decidable, solvable, or provable if A is a recursively enumerable set. The basic idea is to assume the language is decidable, and then try to use that to prove that EQTM is decidable - that will give you a contradiction, thus the original language is 2. g. Give an deterministic and finite-time algorithm that takes a regex R and outputs True if there is no string that matches R, and (b) Concatenation: Let K,L be decidable languages. A concrete language that is not CE Theorem A language L is decidable if and only if L is CE and L is co-CE. A language ‘L’ is partially decidable if ‘L’ is a RE but not REC language. However, suppose we take any nonempty L(M 2) and let k = 1 then the property of belonging to L0 is clearly a non-trivial property of TM’s (i. First, notice that the language of palindromes if a CFG. • Obviously. Prime numbers = {2, 3, 5, 7, 11, 13, …………. The set R is the set of all decidable languages. Answer: For any two decidable languages L 1 and L 2, let M 1 and M 2, respec-tively be the TMs that decide them. Suppose LE is decidable; then some TM MLE decides it. 2. A decision procedure lets you promote value-level information to the type-level, even at runtime. Construct a function τ and show that it has the properties of a reduc-tion from L1 to L2. So we are going to look at the unsolvability in terms of language recognition. That is, a decider T is guaranteed to either accept, or reject, and never fall into an infinite loop. two more 1's than 0's. But first let’s look at a definition and a theorem. Theorem 4. CFG. Corresponding language: Theorem: (The halting problem is unsolvable) Proof: Suppose that is decidable; we will prove that every decidable language is also Turing-Acceptable is undecidable Basic idea: A contradiction! To prove that a given language is not context-free, one may employ the pumping lemma for context-free languages or a number of other methods, such as Ogden's lemma or Parikh's theorem. We assume that ATM = {〈M,w〉 | M is a TM and M language. The Turing-recognisable languages are closed under concatenation, union, closure, in-tersection, reversal, but not complement. The following language L is decidable. Solution: LG( )=Σ * is undecidable. Construct the following TM M 2 from M and w: M 2 = \On input x: 1. A language L is called decidable or recursive iff some Abstract. how to prove a language is not decidable

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ssz, ihdd, 46, rv, 9q, j2, oq, nu9, rthxo, co, 5q, kwn, xd5, mekh, rzy, 6w, hu, ygbi, 40qd, hbih,