If A is ﬁnite, it is decidable because all ﬁnite languages are decidable (just hardwire each of the strings into the TM). Problem 9. What I've been able to pull up online seems concerned about decidability in the context of theorem proving software and artificial intelligence. Prove or disprove each of the following statements. We develop examples of languages that are decidable by algorithms A language is decidable if there is an algorithm (i. 22 A language is decidable iff it is Turing-recognizable and co-Turing-recognizable. • Let L be the language with 0, s, +, ·. , B)A A language ‘L’ is decidable if it is a recursive language. Michael Sipser 1 How to prove Turing decidability of languages Language hierarchy Recognizability A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. Introduction-to-the-Theory-of-Computation-Solutions / chapter4. De ne the function f : ! as f(w) = n 1 if w 2A; 0 if w 62A: Observe that A is a context-free language, so it is also Turing-decidable. A recursive language is a formal language for which there exists a Turing machine that, when presented with any finite input string, halts and accepts if the string is in the language, and halts and rejects otherwise. ) Given an example of a language Lsuch that Lis co-Turing recognizable but its complement is not. y, and testing whether hx;yi 2 D. Run M on w. A language L is decidable if there exists a TM M such that for all strings w: –If w ∈ L,M enters qAccept. , B)A Decidable Languages Consider the following language about DFAs: A dfa = { <M, w> | M is a DFA accepting input w} Is this language decidable? That is, given as input, a DFA and a string, can we decide whether the DFA accepts the string? The answer is yes. You can take the retract in which the language is empty. In this section we prove that it is Theories through axioms The number of sentences that are necessary for defining a theory may be large or infinite. (10 points) Let the language A be the pairs hD;wiwhere D is (an encoding of) a DFA with alphabet and w 2 is a string such that w 2L(D). there exists a TM such that (a) Prove or disprove: The complement of a decidable language is 18 Oct 2018 A language is (Turing) decidable if there exists a Turing machine that decides Are any languages undecidable, and if so, how do we prove it? Prove that Turing-recognizable languages are closed under the prefix operation. Problem 11. Using B’s solution to solve A { I. For a decidable language, for each input string, Theorem: A language is decidable iff both it and its complement are recognizable. We demonstrate how to prove decidability of various algebraic theories in the literature. e. And, if B is decidable, then A must be decidable. Our model is strictly more general than the well-known communicating finite-state machines and allows: (1) individual components to be visibly pushdown automata, which are more suitable for modeling (possibly recursive) programs, (2) the set of words (i. To prove that the set of all Turing-recognizable languages is a proper superset of the set of all Turing-decidable languages, we’ll examine the language A TM. In particular, give an algorithm Posted one year ago is undecidable, i. the pumping lemma for regular languages or the pumping lemma for context-free languages. Based on the Pumping Lemma, we can state some decidability results for context -free languages. If G does not generate w the algorithm doesn’t halt. INF is undecidable. Then that's just the theory of the infinite set, which is recursively axiomatizable and complete and so it's decidable. CS235 Languages and Automata Department of Computer Science Wellesley College Reduction A Tool for Proving Undecidability Wednesday, November 16 and Friday, November 18, 2011 DecidableLanguages Ourpowerisinourabilitytodecide. . Homework 8Solutions 1. Proof: in class Theorem. So a system can be complete and undecidable, right? If T is an axiomatizable theory then no. 3 Some undecidable languages Now that we know DIAG is not Turing recognizable, we can go back and prove that A DTM is not decidable. Let L 1 and L 2 be two semi-decidable languages, and let M 1,M 2 be Turing machines such that L(M 1) = L 1 and L(M 2) = L2. M = “On input <B, w>, where B is a DFA and w is a string: 1. THEOREM 4. Run TM that decide on input 4. To prove that a language is in the class decidable language. Namely, on input w, D' feeds w to the original decider machine D A decidable language • To show that a language is decidable, we have to describe an algorithm that decides it ‣We’ll allow informal descriptions as long as we are confident they can in principle be turned into TMs • Consider ADFA = { M,w ⃒M is a DFA that accepts w } • Algorithm: Check that M is a valid encoding; if not reject. Prove this language is decidable. Let A = f(n,m)j Every n-state machine M either halts in less than m steps on an 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. Construct the DFA that accepts 2. Also, w 2A if and only if f(w) = 1, which is true if and only if f(w) 2B. Prove that ECFG is a decidable language. In other words, there exists an order of all languages. that undecidable language to the given language). If A reduces to B, and B is decidable, then A must also be decidable, since a solution to B provides a solution to A. To prove a certain language is undecidable, we can come up with a reduction that reduce a known undecidable language (A) to our target (B). Universal language A TM = fhM;wijw2L(M)g. Problems and solutions. Hence, the TM that decides is: = "On input where is a regular expression: 1. Proof in two directions: First, if A is decidable, show both A and its complement are Turing-recognizable. 5. 187-188) W e say that problem A reduces (or is reduc ible) to problem B , if we can use a solut ion to B to solv e A (i. cides the language A, then based on the discussion from the previous lecture we conclude that A is decidable by an ordinary DTM, and is therefore 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, … 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. If the simulation ends in an accept state, accept. 2. In order to decide the HOM problem In this paper we prove decidability results of restricted fragments of simultaneous rigid reachability or SRR, that is the nonsymmetrical form of simultaneous rigid E-unification or SREU. Lemma. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). ” This is a computable function because testing whether a string is in a context-free language is decidable. But N behaves the opposite of Mon input hii, contrary to the assumption that D= L(M). Let R be its recognizer, clearly its running time is in nite =) hRi62A =) A 6= fhMig. A 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. We say that a language is co-Turing-recognizable if it is the complement of a Turing-recognizable language. Proof: (By contradiction) Suppose that context-free languages are closed under complementation. 8. A decidable language • To show that a language is decidable, we have to describe an algorithm that decides it ‣We’ll allow informal descriptions as long as we are confident they can in principle be turned into TMs • Consider ADFA = { M,w ⃒M is a DFA that accepts w } • Algorithm: Check that M is a valid encoding; if not reject. Set enumerability Given a set S, the characteristic function of S is de ned as 1. Although it might take a staggeringly long time, M will eventually accept or reject w. sets. Proof. Solving B 2. Theorem 16. A TM is undecidable. Prove that the halting problem is recognizable (Hint: Similar to the theorem on Slide 4. Instead, it is common to define a theory through a set of language 2BQP and let g be a language 2BQPf, a language decidable by a BQP device with access to f. On receiving input w, M runs E to enumerate all strings in A in lexicographic order until some string lexicographically after w appears. Proving the non-existence of algorithms for computational problems can be very The class R of recursive (=decidable) languages over an alphabet Σ. , if B is decidable/solv able, so is A . Lecture 17: Proving Undecidability. The atomic formulas over this language are equality over string terms (word equations), linear inequal- 3/17/2016 Using reductions to prove undecidability We want to prove that language L is undecidable. Let L and L be decidable languages. 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. A many-one reduction: A m Bif exists a computable function fsuch that 8x2 A, x2A f(x) 2B. If M1 accepts, accept. We use a similar technique to prove that there exists some language that Theorem 10 A language is decidable if and only if both it and its complement are 1 Nov 2016 A language is decidable if there is a TM for which this is the case. If accepts reject; if rejects accept. Recursively enumerable language(RE) – A language ‘L’ is said to be a recursively enumerable language if there exists a Turing machine which will accept (and therefore halt) for all the input strings which are in ‘L’ but may or may not halt for all input strings which are not in ‘L’. (Recall In mathematics, logic and computer science, a formal language is called recursive if it is a For example one may speak of languages decidable on a non-deterministic Turing machine. g. Decidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. Convert NFA B to an equivalent DFA C using the procedure for this conversion given in Theorem “subset construction”. Then if and are context-free languages, so are and . Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. But we can also prove that there is a constant c > 0 such that every Turing Machine that decides PR requires time at least 22n c. of the problem has answer Yes or No, then the problem is said to be decidable. For any language, if it is decidable, then it is also recognizable. Second, we show that the satisfiability problem for word equations in solved form that are regular, The HOM problem is decidable Guillem Godoy Omer Gimenez Lander Ramos Carme Alvarez June 26, 2009 Abstract We close a rmatively a question which has been open for 35 years: decidab Abstract. Show that the collection of Turing-recognizable languages is closed under the oper- Prove that a language L is decidable if and only if some enumerator . 1 What is reducibility? Reading: Sipser 5 (pp. A finite language is always decidable. Prove that the halting problem is undecidable (Theorem 5. (e)Recognizable sets are closed under complement. Obtaining an actual description of a DTM that decides the lan- 5. For TOTAL TM, assume that it is the range of a computable function (see the previous problem) and then diagonalize to obtain a contradiction. L ∈ R iff L is decidable Undecidable Languages The Question: Are there languages that are not decidable by any Turing machine (TM)? i. Problem10. 15. Thus, it is Turing-recognizable, or semi-decidable. To prove INF is undecidable we can either refer to Rice’s theorem (arguing that INF is a non-trivial property), or give a direct reduction, e. – Recursively The classes of Turing- recognizable and Turing-decidable languages We'll prove that they cannot. A Turing machine is needed to know though whether a word w is in the Language. Now suppose a language L and its complement are recognizable. 7. Thus, f is a computable function. 840 Theory of Computation (Fall 2013), taught by Prof. Proposition 15. 6 (page 142). (Hint: it might help to prove a rigorous version of the statement: If a binary measurement on a quantum state outputs 0 with high probability, then the post-measurement state on output 0 has high overlap with the Using proposition 3. Prove that a language L over Sigma is decidable if and only if there is a TM E that enumerates L in such a way that the strings in L are output by E in length-increasing fashion. a). Or: Construct a (mapping) reduction from another language already known to be non-Turing-recognizable to the given language. if a language doesn’t satisfy pumping lemma, then we can definitely say that it is not regular, but if it satisfies, then the language may or may not be regular. Nov 08, 2016 · Reducibility: Used to prove some language is undecidable. Prove that the language of M is decidable . Construct the DFA that accepts 3. Otherwise, the problem is undecidable. Prove that the language f<M;w;q>jMis a Turing machine which visits state qduring its We will prove that the languages are closed by creating the appropriate grammars. halting on an accepting state) or input do not belong to the language (rejection i. Every language is countable. CS4313/5353 Theory ofComputation Decidable Languages –2 ProblemsConcerning RegularLanguages ADFA = Languages decidable by weak models of computation often have certain necessary characteristics, e. 00 C 1983, Elsevier Science Publishers B. IScan the input string repeatedly. Suppose we have two context-free languages, represented by grammars with start symbols and respectively. So if A is undecidable, then B must be undecidable. Hints: Recall that a language is decidable if and only if both it and its complement are semi-decidable. Then: To get the union, add the rule 10 Reducibility 10. decidable We know that a language may be semi-decidable but not decidable. Your language L is indeed undecidable. The diagonalization language, which we The set R is the set of all decidable languages. The absence of symmetry enforces us to use different methods, than the ones that have been successful in the context of SREU (for example word equations). The easiest is to use deﬁnition 3. That is, Dis a decider for the language: INFINITE CFG= {<G>| Gis a CFG and L(G) is infinite}. A language L is decidable if and only if L is decidable. If A is inﬁnite, a TM M that decides A operates as follows. it is deﬁned by a regular expression, and A is one of our canonical examples of a non-regular language. CFG is a decidable language Proof ideas: For a CFG G and a string w: I First idea: Go through all derivations generated by G checking whether any is a derivation of w. Bucher, J. A language L is decidable if Review: An instance of the halting problem asks whether Turning machine N halts on input y. Proof languages Lcontaining other kinds of entities as, for example, numbers. The TM can only move from q 0 to q 4, the reject state on encountering a 1. Proof: Let M1 be a TM that decides a language L1, and M2 be a TM that decides L2. Problem I. ) and reduce Ato B. Lemma: It is decidable whether or not a given string belongs The set of all languages over the alphabet {0, 1} is uncountable. Prove that there exists a decidable language C such that A C and B C. 39. A ≤ B, means A is computationally easier than B. Decidable and recognizable languages. Hence, all computation paths will reject. The definitions of mapping reduction and mapping reducibility. Prove that its complement is undecidable. Prove Disprove L Turing Decidable Turing Recognizable Language Draw Turing Machine Q13470117 January 5, 2020 / 0 Comments / in Academic / by admin Prove or disprove that L is a Turing-Decidable orTuring-Recognizable Language. Language Decidability - A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. Answer: Deﬁne the language as C= {hM,Ri | M is a DFA and Ris a regular expression with L(M) = L(R)}. (10 points) Assume that the language L is decidable. Prove that ALLDFA is decidable. Usually, not every (possible) string is representing an object. Jun 01, 2016 · How to Prove whether a regular language is finite is decidable Or it is decidable to check if a given regular language is finite. PA proves all decidable things so you can define a truth definition for formulas in that retract. If A m B and B is a regular language, does that imply that A is a regular language? Answer: No. (a) The complement L is decidable. The reason is that language can be expressed using regular operators as . Prove that if a language A is mapping reducible to a language B and B is decidable, then A is decidable (Theorem 5. To prove a language is decidable, we can show how to construct a TM that decides it. Recursively enumerable language(RE) – A language ‘L’ is said to be a recursively enumerable language if there exists a Turing machine which will accept (and therefore halt) for all the input strings which are in Jun 01, 2016 · An NFA accepts an infinite language if and only if there is a (directed) path from the initial to an accepting state which contains at least one state more than once. Consider any input string starting with a 1. To prove that a given language is non-Turing-recognizable: Either do both of these: Prove that its complement is Turing-recognizable. This states that a language is decidable if some Turing machine decides it. A language is decidable. –If w ∉ L, M enters qReject. 17 Nov 2014 Define a language L Turing-Acceptable iff. To prove that a given language is Turing-recognizable: Construct an algorithm that accepts exactly those strings that are in the language. We construct a TM T, deciding A dfa. Apr 25, 2008 · Deciding weak bisimilarity is harder since the usual decomposition property which holds for strong bisimilarity fails. 2) This language could be decided by a DTM similar to U deﬁned above, but where it cuts the simulation off after t steps if M has not accepted w. Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. Let us say the language accepted by the given Turing machine M is L has then given an input if we can decide in finite time that input belongs to the language (acceptance i. argument) Paul Goldberg Intro to Foundations of CS; slides 3, 2017-18 25/42 Homework 10 Solutions 1. The language fn 2N : nis primegis decidable since a program that checks whether any integer from 2 to p nis a factor can tell whether or not nis prime. The HOM problem consists in determining, given a tree homomorphism H and a regular tree language L represented by a tree automaton, whether H(L) is regular. From Cambridge English Corpus The emphasis is on achieving maximum expressiveness whilst maintaining sound, decidable and understandable typechecking. If M accepts, accept; otherwise, reject. 23). 7. On each computation path, the TM does one of the following : A language is Turing-recognizable if there exists a Turing machine which halts in an accepting state i its input is in the language. Thus, this language is not decidable but it is recognizable. True. Answer: If D exists, we can construct a TM M such that we search each possible string. 3 For example one may speak of languages decidable on a non-deterministic Turing machine. Consider the decision problem of testing whether a DFA and a regular expression are equivalent. 1 in text) Suppose AH is decidable there’s a decider MH for AH Then, we can construct a decider DTM for ATM: On input <M,w>, run MH on <M,w>. Let A and B be Turing-recognizable languages such that A S B = . Let A How to Prove Monad Laws are SN by SOL Step 2. In this paper, we present a tableau method to decide weak bisimilarity of totally normed BPP. Run M on input <C, w >. For a correct proof, need a convincing argument that the TM always eventually accepts or rejects any input. L. For every decidable subobject D of N, there is an n ∈ N such that n ∈ A 0 ∩D or n ∈ A 1 –D. A language L is Enumerable if there is a Turing Machine that lists the strings w∈L in any order. This shows L is decidable since for every L with a de-enumerator, we have a Turing machine that decides L. Since L is Turing decidable, there is a Turing decider machine D (one that halts on all inputs) that recognizes L. Decidable: Booleans and decision procedures. Theorem. On the other hand, it is easy to construct a Turing machine D' that recognizes L'. We prove several decidability and undecidability results for the satisability /validity prob-lem of formulas over a language of nite-length strings and integers (interpreted as lengths of strings). ¼Equivalent language: AH = { <M,w> | TM M halts on input w} ¼Need to show AH is undecidable ¼We know ATM = {<M,w> | TM M accepts w} is undecidable)Show ATM is reducible to AH (Theorem 5. on suﬃciently large inputs of size n. Prove that neither TOTAL TM nor its complement is semi-decidable. Since there are in nitely many derivations this idea does not work. Other languages based on numbers and arithmetic One can imagine many other languages based on the natural numbers, arithmetic, and so on. Lecture 17: Proving Undecidability 4 If a language L is decidable, then its complement L' is also decidable If a language is decidable, then there is an enumerator for it. So if M accepts w, M2 accepts not only 0^n1^n kind of inputs but Sigma. 1) Prove that the halting problem is recognizable (Hint: Similar to the theorem on Slide 4. 0304-3975/83/S3. A dfa is decidable. A TM is recognizable but its complement is not recognizable. Invoke the SOL’s command sn in GHCi *SOL> sn monad sigm Found constructors: return Checking type order >>OK Checking positivity of constructors >>OK Checking function dependency >>OK Checking (unitL) return(X) >>=y. The methods … 23 Oct 2014 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 13 Apr 2017 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. A decision problem P is decidable if the language L of all yes instances to P is decidable. If it ends in a nonaccepting state, reject. In this case we have to represent these objects as strings. Aug 08, 2005 · Decidable: A set of sentences R is decidable if the set of sentences of its language that are consequences of R is recursive. quire that a Turing machine halts even if it does not accept. Basic technique for proving a language is (semi)decidable is reduction Based on the following principle: { Have problem Athat needs to be solved { If there exists a problem B, such that B’s solution will enable the solution to A, you can solve Aby 1. " Note: if accepts, it means that , i. It must There are two equivalent major definitions for the concept of a recursive (also decidable) language: 2. By definition , all REC languages are also RE languages but not all RE language is decidable: S DTM = ˆ hM,w,ti: M is a DTM, w is a string, t 2N, and M accepts w within t steps ˙. ) * (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. YES NO Run with input reject accept reject Decider for Input string Decider for and halt halts and rejects and halt and halt Contradiction!!!! END OF PROOF Therefore, is decidable But there is a Turing-Acceptable language which is undecidable Since is chosen arbitrarily, every Turing-Acceptable language is decidable *. To prove: Every context free language A is decidable ; Incorrect Proof outline: Since A is CF, there is a PDA that recognizes A. Let T be the theory in L whose only non-logical axiom is sx · sy = x· y +(x +sy). All decidable languages are recursive languages and vice-versa. Such characteristics are useful for determining when a language cannot be decided by that weak model of computation. (North- llolland) 23- W. The complement of a language is the language consisting of all strings that are not in the language. Simulate M1 on w. If A reduces to B, and A is undecidable, then B must also be undecidable. 1 in text) ¼Suppose AH is decidable ⇒there’s a decider MH for AH ¼Then, we can construct a decider DTM for ATM: On input <M,w>, run MH on <M,w>. N = “on input <B,w>, where B is a NFA and w is a string: 1. Show that for any alphabet there is a language L over such that neither L nor If a language is both semi-decidable and co-semi-decidable, then it is decidable. 22). Moreover, if a language is decidable, then so is its complement, and hence that complement is recognizable. How to Prove Undecidability or Non-Turing-Recognizability in This Course To prove that a given language is undecidable: Construct a (mapping) reduction from another language already known to be undecidable to the given language. 2. Next we create a matrix in which we list all strings across the columns, X 1;X 2;X Lecture 7: Other undecidable languages, Rice’s theorem, and Reductions Valentine Kabanets October 4, 2016 1 Semi-decidable vs. V. INF is not semi-decidable. [10 marks] 4. right now, easy typically means decidable right now, hard typically means undecidable One option: prove from scratch that the problem is easy (decidable), or prove from scratch that the problem is hard (undecidable) (e. The class of semi-decidable languages is closed under union and intersection operations. IScan the input string hB;wi, determining whether the input constitutes a valid DFA and string w. A Turing machine Mrecognizes a language Lif and only if: 1. The problem is known to be undecidable (but semidecidable). Simulate B on input w. That is, prove the following two statements. Every decidable language is Turing-Acceptable. , the language consisting of all Turing machine descriptions that are POLITICIANs is undecidable. Decidability Techniques for Proofs Miscellaneous - 1 Problems on Language Classes and TM Miscellaneous - 2. This is actually a really standard example in theory of computation as well. yes instances to P is not decidable or a language is undecidable if it is not one can become very quick in coming up with proofs for these problems on the spot. The complement of INF is not semi-decidable. In this case we are only interested in the set of all well formed strings instead of itself. We show that it is decidable whether or not a relation on the reals deﬁnable in image of a context-free language. that Ddecides if the language of a CFG is infinite. Similarly for decidability. 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. c) If A ≤m B and B is a regular language then A is a regular language. a Turing Machine decider) to. 3. Therefore one way of showing that a language is decidable is by describing a Turing machine that accepts it. That is, show that if L1 and L2 are decidable languages, then L1 intersection L2 is a decidable language. The complement Lc = nLof a language L, is de ned in Lbe a nite relational language. This is decidable in polynomial time. First of all, rename all the terminal symbols in the second grammar so that they don't conflict with those in the first. Turing Decidable Languages are both Turing Rec and Turing Co-Recognizable. In this paper we prove that the isomorphism problem in this setting is decidable, even when the assumption of non-singularity is removed. Consider a partially decidable language L. (only if): Follows from the previous lemma and the fact that every decidable language is Turing-recognizable. (15. , languages) of messages on queues to form a visibly pushdown language, which permits Every context free language is decidable - A Wrong Approach. Give an example of a DFA Astallion that in in ALLDFA. Decidable problem: there is an algorithm for solving it. ). tex Find file Copy path ryandougherty Added Table of Contents, changed document format 3008db2 Jul 12, 2016 There is a well established theorem to identify if a language is regular or not, based on Pigeon Hole Principle, called as Pumping Lemma. Undecidability; proof by diagonalization and getting the paradox. The following is a mapping function from A to B: “On input w: 1. 6. Here are just a few examples: C = Aug 20, 2019 · 1. iii) The emptiness problem of nonzero automata is in np ∩ co-np. Proposition 3. To prove that Bis undecidable, (not semi-decidable, not co-semi-decidable) pick Awhich is undecidable (not semi, not co-semi. 1 if w 2A; 0 if w 62A: Observe that A is a context-free language, so it is also Turing-decidable. The set R is the set of all decidable languages. , A TM < INF as follows. Then show that g 2BQP. In computability theory, a set of natural numbers is called recursive, computable or decidable if there is an algorithm which takes a number as input, terminates after a finite amount of time (possibly depending on the given number) and correctly decides whether the number belongs to the set or not. More precisely it is decidable - regular languages are closed under complement and intersection, - given a context free language G and a regular language A, the language L(G) intersection A is also a context free language, and - emptiness is decidable for context free languages. Consider the language INF = fhMijL(M) is in niteg. Convert B into an equivalent DFA C using the procedure for this conversion given in Theorem 1. Therefore, whenever an ambiguity is possible, the synonym for "recursive language" used is Turing-decidable language, rather than simply decidable. Hirshfeld proposed the notion of bisimulation tree to prove that weak bisimulation is decidable for totally normed BPA and BPP processes. Otherwise output 10. (4 pts) recognizable are separated by some decidable language. This seems sensible to me, for the same reason that there are truths about the world that we will never be able to prove, like the existence of a flower in a forest fire. This known undecidable language can be any language for which undecidability has been to the language, the computation of the TM either rejects or goes on forever. Prove that some decidable language D is not decided by any decider M i whose description appears in A. If a Language is Not Turing Decidable, either it, or it's complement, must be not Recognizable. If w is in A then output 01. Languages recognized by a TM are called recognizable. if there is a Every decidable language is Turing- Acceptable. Solution: We may assume that Ais in nite since there are in nitely many decidable languages. For that we rely on ii): the states of the Prove that if a language L and its complement L are both recognizable, then L is decidable (Hint: the idea is, in fact, already given in the proof of Theorem 4. A set A is countable if either it is finite or it has the same size as the set of integers {1, 2, 3, … }. Recall that: A language A is decidable, if there is a Turing machine M (decider) that accepts the language A and halts on every input string Decision On Halt: Turing Machine M YES Accept Input Decider for A string NO Reject it is in the set, and a decidable language is one that has a Turing machine that will always halt and give a yes or no answer as to whether its input is in the language. Proof: Again, we construct a Turing machine that decides this language. De ne the function f : ! as f(w) =. GOAL: To prove that H does not exist (proof by contradiction). 14 Theorem 5. Fundamentally, all Turing machines accept a Language L that is defined by the Turing machine as all input strings that the Turing machine accepts. L is decidable if and only if both L and L are Turing-recognizable. 1. Equivalently, a formal language is recursive if there exists a total Turing machine (a Turing machine that halts for every given input) that, when given a finite sequence of symbols as input, accepts it if it belongs to the language and rejects it otherwise. ” Proof: To represent B, simply list its components Q, Σ, δ, q 0, and F on the tape. 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. In other words, a language is decidable exactly when both it and its complement are Turing-recognizable. We have a choice as to how to represent relations: as an inductive data type of evidence that the relation holds, or as a function that computes whether the relation holds. Decidable and Semi-Decidable Languages Decidable A language L is Decidable if for every string w, there is a Turing Machine M that correctly Decides whether w∈L • M Halts and Accepts if w∈L • M Halts and Rejects if w∉L Semi-Decidable A language L is Semi-Decidable if for every string w, there is a Turing Machine M that Semi-Decides We want to prove is the following reduction: Given a general grammar G , find a context-free language D , such that L(G) = ∅ if and only if D = Σ * How to decide if a language is In R or RE or CoRE2019 Community Moderator ElectionIs a CFG using a nil What is the purpose of a disclaimer like "this is not legal advice"? Idiom for feeling after taking risk and someone else being rewarded Let Abe a Turing recognizable language consisting of descriptions hMiof Turing machines M that are all deciders. A is decidable A is also Turing-recognizable A is decidable A's complement is decidable A's complement is Turing-recognizable. we prove that the 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. 3. a) Every recognizable language can be recognized by a TM that either accepts or loops, but never rejects. Prove that for a Turing machine M the halting problem language { (M, w) | w is in H(M) } 27 Oct 2011 Proof. There are three methods you may use to prove this is true. • A Turing Using mapping reductions, prove that S is neither r. Homework: Prove that the language H of all halting algorithms is undecidable, 18 Dec 2014 Exercise 7. As an example, there is a number of open questions concerning connections of various classes of pure languages to classes in the Chomsky hierarchy. 8s2L, MAccepts s We haven’t actually shown that there’s a meaningful distinction here. If B were not undecidable, than we could use the solution to B to decide A (a contradiction since A is undecidable). 4. D exists such that C = fx j 9y(hx;yi 2 D)g. Language is Turing recognizable if some Turing machine recognizes it •Also called “recursively enumerable” Machine that halts on all inputs is a decider. K[y]] [is positive in return(X)] [is acc in K[y]] Lecture 16: The Universal Language 1. • If the halting problem were decidable, then the universal language would also be decidable • Reducibility says nothing about solving either of the problems alone; they just have this connection • We know from other sources that the universal language is not decidable • When problem A is reducible to problem B, solving A cannot be Then, D recognizes the same language as M, and it is semi-decidable because it may loop on some inputs. Decidable and Semi-decidable Languages (Score: _____ out of 20 points) Let Sigma be an alphabet. But if M does not accept w, M2 will accept just 0^n1^n strings. p = n ∪ ERR, where ERR is the easy to decide language: ERR = { x ∈ { 0 On the other hand, there exists decidable languages, which. S(x) = ˆ 1 if x 2S 0 if x 2=S (1) If the characteristic function for a set S istotal and computable, S is decidable (or recursive) If 1. The class of all recursive languages is often called R, although this name is also used for the class RP. Is T decidable? • Is there a theory in which the Godel sentence expressing inconsistency is true, and yet which is consistent? Why? • Use the Completeness Theorem to prove the existence of non-Archimedean ﬁelds. 21 there is an enumerator Efor the language A. Prove that C is Turing-recognizable i a decidable language D exists such that C = fx: 9y; hx;yi2Dg. That problem would be called decidable. Languages decided by a TM are called decidable. 2 A non-Turing-recognizable language We will now deﬁne a language and prove it is undecidable. • If the halting problem were decidable, then the universal language would also be decidable • Reducibility says nothing about solving either of the problems alone; they just have this connection • We know from other sources that the universal language is not decidable • When problem A is reducible to problem B, solving A cannot be Can a vampire attack twice with their claws using Multiattack? Can you really stack all of this on an Opportunity Attack? Is it possible Lemma: The context-free languages are not closed under complementation. 5 The set of all languages is uncountable. , B)A T decides a language L if T recognizes L, and halts in all inputs. Proof: We prove this theorem by contradiction. Hence the property is non-trivial. Here we explore the relation between these choices. On input x, M simulates (N, y) for length(x) steps. Proof: Surely, a decidable language is recognizable. (Hint: You may nd it helpful to consider an enumerator for A. Recursive languages are also called decidable . 1 (Decidable Languages). These are all decidable languages (once again, for any reasonable en-coding scheme). By Thm 3. The language accepted by the machine are all strings starting with a 0. Idea: Use a proof by contradiction. But pumping lemma is a negativity test, i. For some undecidable language L1, and decidable language L2, how do I prove that the union, L1 U L2 is undecidable? I've already proved that for some undecidable language L1, and finite language L2, that the union L1 U L2 is undecidable, which can be seen here: Proof Basic technique for proving a language is (semi)decidable is reduction Based on the following principle: { Have problem Athat needs to be solved { If there exists a problem B, such that B’s solution will enable the solution to A, you can solve Aby 1. False. (20 points) Show that the set of decidable languages is closed under intersection. Examples of how to use “decidable” in a sentence from the Cambridge Dictionary Labs of the definitional equalities in object languages. Theorem: The class of decidable languages is closed under union. TM is not decidable. Prove that a one-tape Turing machine which cannot overwrite their input are equivalent to a Finite automata. Is there a Turing machine that takes a language as input and decides/semi-decides if it is a decidable language? Comments + answer say trivially the answer is yes; however, I'm wondering here would it be possible without the use of excluded middle to construct such a machine. Prove the following properties. Prove that there is no onto function g: N!P(N). To prove iii) we provide a reduction of the emptiness problem to the computation of the winner of a parity game called the jumping game. What Decidable Means. De nition 2. It is clearly a property of the language of M. Proof: N = “On input <B, w >, where B is a NFA and w is a string: 1. If such y exists, accept. A is decidable A is also Turing-recognizable are decidable languages Idea: Use previous machine M as a subroutine. Proof To show Proof To prove that LC is not decidable, we assume that it is decidable by the TM MC and. Two sets A and B are the same size if there is one-to-one correspondence (one-to-one, onto mapping) from A to B. Recursively enumerable language: there is a TM that halts if the string is accepted. Theorem 8. Assume for a contradiction that D= L(M) for some hMiin A. TEDx Talks Recommended for you Alin Tomescu, 6. Since M is a dfa, we Proofs of Decidability. BTW, does D exist? Theorem 3 : is a decidable language. By Rice’s theorem, we know that A is undecidable. Countable and Uncountable Sets. n. Theorem 3 : is a decidable language. Suppose to the contrary that L is decidable. That question can be answered by a machine in finite time, but what does it have to do about decidability? the obvious choice is an assumed decider for a given decidable language. Now consider an input string starting with a 0. Then hMiis the i-th machine description printed by Efor some i. Hagauer This paper solves one of these problems: We prove that it is decidable whether a reular Languages decidable by weak models of computation often have certain necessary characteristics, e. W e may use reducibility to pro ve undecidability as follo ws: Assume we wish to pro ve problem B to be undecidable Aug 01, 2013 · The HOM Problem Is Decidable ´ GUILLEM GODOY and OMER GIMENEZ, Universitat Polit` cnica de Catalunya e We close affirmatively a question that has been open for long time: decidability of the HOM problem. Prove that some decidable language Dis not decided by any decider Msuch that hMiis in A. Construct a TM M that decides L1 ∪L2 in the following way: M = “On input w: 1. b) If a language is decidable, then every proper subset of that language is decidable. Assume M is a TM whose program only allows the tape head to move right or stay stationary, but that it never moves left. We will construct Turing machines M L 1∪L 2 and M L 1∩L 2 accepting union and intersection of L 1 and L 2, respectively. ” Hence, A NFA is decidable. Call this PDA P. Jan 22, 2015 · Why I read a book a day (and why you should too): the law of 33% | Tai Lopez | TEDxUBIWiltz - Duration: 18:26. Prove that C is Turing-recognizable if and only if a decidable language. A language ‘L’ is decidable if it is a recursive language. Therefore, you may provide a decider Turing machine M such that L(M) = L to prove L is decidable. We will prove the following: 1. Suppose the set of all languages is countable. That is, if is a context-free language, it it not always true that is also. A decider that recognizes language L is said to decide language L Language is Turing decidable, or just decidable, if some Turing machine decides it 2 Example non-halting machine A language is decidable, if there is a Turing machine (decider) that accepts the language and halts on every input string A M A Turing Machine Input string Accept Reject M Decider for A Decision On Halt: Decidable Languages YES NO Let Dbe the language it decides. universal language, that can be accepted by a Turing machine but is still undecidable. K[y] => K[X] (meta K)[is acc in return(X),y. 2 The language LET is not decidable. A TM is recognizable but not decidable. This shows immediately that L is Turing recognizable. Express this problem as a language and show that it is decidable. Then Mis nite automaton presentable (for short, FA-presentable) if if the elements of the domain can be represented by( nite) words in a regular language D ˆA over some nitealphabet A in such a way that for each relation R of L, we have some nite automatonwhich recognizes the graph of R. halting on a rejecting state). For example, de ne the languages A = f0n1n jn 0gand B = f1g, both over the alphabet = f0;1g. We can prove that PR is decidable. Lexicographically Enumerable A language L is Lexicographically Enumerable if there is a Turing Machine that lists the strings w∈L in Lexicographic order. (f)Decidable sets are closed under complement. (d)There is a recognizable but not decidable language. Give lexicographic-order enumerator: is T-decidable ⇔∃ an enumerator such that prints all of the strings in in lexicographic order. Let A,B be two languages such that A \ B = ∅ and A and B are recognizable. But you are talking about a machine that determines whether certain properties hold for these two words. Therefore, whenever an ambiguity is possible, the THEOREM 5. Prove that the concatenation L 1L 2 = fab; a2L 1;b2L 2gof (partially) decidable languages is (partially) decidable. We just need to Guide To Classifying Languages Claim: L is decidable. 2 Since Sigma* (Sigma = alphabet set) is a regular language, for R to decide whether M2 accepts a regular language it must consider all possible inputs (Sigma ), including 0^n1^n and other nonregular languages. The Turing machine and the Language are therefore in fact equivilant. Costas Busch - We will prove that there is a language : is not Turing- Definition: A language is called decidable if there exists a method - any Proof: Let L and M be languages that are decided by algorithms A and B respectively. The language ADTM is undecidable. How can you prove a language is decidable? 3. Let C be a language. Prove that language fw 1w 2 w k; k2N;w 1;w 2;:::;w k 2Lg We show that if word equations can be converted to a solved form, a form relevant in practice, then the satisfiability problem for Boolean combination of word equations and length constraints is decidable. nor co-r. 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. The language is decidable. Since for every language L ∈ NP there is a machine that decides L in time 2nd, we see that PR ∈/ NP. To show that M' is a decider, we will prove that it always halts. For the ﬁrst two languages, we assume that ha,b,ciis an encoding of three natural numbers a, b, and c, and for the third language hniis the encoding of a natural number n. L ∈ R iff L is decidable Equivalent language: AH = { <M,w> | TM M halts on input w} Need to show AH is undecidable We know ATM = {<M,w> | TM M accepts w} is undecidable Show ATM is reducible to AH (Theorem 5. If not, then reject. , ,. A language is decidable, if there is a Turing machine (decider) that accepts the language and halts on every input string A M A Turing Machine Input string Accept Reject M Decider for A Decision On Halt: Decidable Languages YES NO Undecidable Problems (unsolvable problems) Decidable Languages. 8. 3 we can prove the statement which corresponds to the theorem in recursion theory on the existence of disjoint, inseparable r. 7 Apr 2017 There are many regular languages which are not finite; [code ]L should carefully prove the theorem that every regular language is decidable). Deﬁne a new DTM K as follows. There must therefore exist a DTM T that decides A DTM. dream up a diag. Let A i = {n | φ n (n) = i}, for 1 = 0, 1. ii) A nonzero automaton with F∀ = Q is nonempty if and only if its language contains a regular tree of size |Q|. Assume toward contradiction that A DTM is decidable. Recursive language: there is a TM, correspond-ing to the concept of algorithm, that halts eventually, whether it accepts or not. Show that there exists a decidable language C such that B C A. is a decidable language Idea: Present a TM M that decides A DFA. That is, a decider T is guaranteed to either accept, or reject, and never fall into an infinite loop. There must exist a mapping from all strings to natural numbers. (Hint: Look at the proof for EDFA to get an idea. how to prove a language is decidable