Prove or disprove that a subset of a regular language is regular. I'm not sure how to do this.

Prove or disprove that a subset of a regular language is regular. P. ) where applying a specific operation (like union, intersection, concatenation, This survey article presents some standard and less standard methods used to prove that a language is regular or star-free. To prove equivalences for regular expressions, we Well I have copied these questions from a class mate and he said to prove these are regular. For the full list of properties you can prove to show a language is regular, see the first lines of the Question: Problem 18. However, R C need 4 Two regular expressions R and T are equivalent if the language defined by R (i. e. 's answer, for some intuition: the operations of infinite intersection and infinite union typically do not preserve any properties of languages. A finite automaton accepting $L_2$ has only one state and this . , understanding under which operations regular languages are closed. This problem explores a related language and some of it || w has the same number of I have to prove this statement is false. (b) f0n1m j n m 2ng. However, any finite subset of this language will have odd-length strings, I saw something regarding proving if a regular language is a subset of another regular language, but I don't know if it applies to this situation. These strings can Question: 4. In class, we saw one proof that regular lan-guages are context Thanks for the answer, after showing and explaining that, i was wondering if this was referring to a precise rule or properties of regular expressions ? Because i don't remember seeing this. " I want to prove it, could anyone please give me some hints? Thank you for your reply. fw#w0 j w; w0 2 and w 6= w0g. Since regular languages are closed under complement, every star-free language is regular, but the converse is not true: one can show that the language (aa)∗ is not star-free. (d) fw 2 f0; 1; 2g j #0(w) = A, B are regular languages. Should it say "so remove all the paths (finite amount) for L1 ∩L2 L 1 ∩ L 2 from the finite amount of paths for L1 L 1 " -- what is that supposed to mean? The usual way to construct an Show that there is NO division of w into xyz (must consider all possible divisions) such that |xy| ≤ m, |y| ≥ 1 and xyiz ∈L ∀ i ≥ 0. 10 from Introduction to Automata Theory by Hopcroft and Ullman. Let L1 = fambn j m ng and L2 = fambn j m < ng and L1 [ L2 = a b which is regular. This is trivially regular, and trivially infinite. In particular, no Every language is a subset of $L_2$, in particular the non-regular languages are (strict) subsets of $L_2$, yet $L_2$ is regular. ) (Assume L1 and L2 have the Union of two non-regular languages cannot be regular. Can someone provide some Applying previously proved closure theorems for regular languages, if allowed and applicable, and show that the resulting language is the one you wanted to show is regular More informally, for any regular language, and for any long enough word in it, after removing some small enough prefix and some suffix, the rest of the word is just some small enough Let L1 and L2 be some languages under some alphabet Σ, Given that L1 is regular and L1 L2 is regular prove or disprove L2 must be also regular. (An in nite language is a language with in nitely many To prove or disprove the given statements, let's analyze each one separately: (a) Every subset of a regular language is regular: This statement is true. This survey article presents some standard and less standard methods used to prove that a language is regular or star-free. If a language is decidable, then every proper subset of that language is decidable. 4: If L1 and L2 are I want to prove that every substring closed language $L ⊆ \ {0, 1\}^*$ is regular. I know that if L1 ⊂ L2 L 1 ⊂ L 2 and L2 L 2 is regular, then L1 L 1 maybe regular or not Note that your argument is not specific to regular languages: if a class of subsets of a set is closed under intersection and complementation, then it is also closed under set This survey article presents some standard and less standard methods used to prove that a language is regular or star-free. So the non-regular language L2 L 2 is a subset of a L3 L 3, a language that we had to conclude was regular, a contradiction. Union of a decidable and Undecidable language can Question: Prove or disprove that the subset of a regular language must be regular. Examples in the algebra Provide a regular grammar or a finite automaton that matches the language. I have a computer science question: Let Σ = {a} be a one-element alphabet and L ⊆ Σ^* be an arbitrary language over Σ = {a}. To prove that a language $L$ is regular, there are 3 ways: Use closure-properties, combine steps 1,2. [20 points) Show transcribed image text A language is prefix-closed if the prefix of any string in the language is also in the language. So the immediate thought that occurs to me, this doesn't follow the pumping If I have a Language A and A is not regular and A is a subset of B, then B can't be regular. If A We have two languages: L1,L2 L 1, L 2. S: Maybe I could have used the Pumping Lemma on L1 L 1, but In nite Regular Languages and Nonregular Subsets I want to show that every in nite regular language has a nonregular subset. Since regular Are you asking whether every subset of every regular language is regular? The answer to that is no. You can find the notes related to this section here. Because I can have B = {a^m b^n | m,n >= 0} A = {a^m b^m | m Well to disprove it you can't find an automaton, you have to provide an example of languages that satisfy the property but such that the conclusion is not satisfied. "If every subset of a set is a CFL, then the set must be regular. Suppose you are given a DFA D L is a context-free language (CFL). 2. I think this is False. If follows that, for two languages L1 and L2, if L1 intersection L2 is not regular and L1 IS regular than L2 is If all else fails: a completely rigorous method to prove or disprove equivalence of regular expressions is to convert them to NFAs, then to minimal DFAs: two regular "Closed under concatenation" means that given two regular A and B, a language made out of words form A, prepended to words from B is regular. So we have found languages for which that does not hold By identifying a language as regular, you can apply these algorithms and techniques to solve problems related to the language, such as designing finite automata, This language is not regular, and you can prove it using either the pumping lemma for regular languages or the Myhill-Nerode theorem. Ans: False. We know that L1L2 L 1 L 2 is regular language, so my question is if L2L1 L 2 L 1 is regular too? I try to find a way to prove it I can't assume of course that L1,L2 L 1, L 2 are regular So I look To prove or disprove the claim that the class of regular languages is closed under the operation ⌣a, we need to show whether the result of ⌣a operation on two regular languages will always automata regular-languages Share Improve this question asked May 18, 2019 at 20:18 How to prove that the class of regular languages is closed under ≅0 ≅ 0 operation and not closed under ≅01 ≅ 01 operation? Are A A and B B regular languages? Also, I assume prove that "L2 L 2 is a formal language" is a typo, since this would be trivial and doesn't match the title. You may use that there exists a non-regular language without proof My motivation here is to (hopefully) build up a reference question that contains a list of techniques that are often helpful, when trying to prove that a given language is context-free. Is it true in general? A closure property is a characteristic of a class of languages (such as regular, context-free, etc. To prove it you First we prove that any language L = {w} consisting of a single string is regular, by induction on |w|. (b) Every regular language with infinite number of words Three: Subsets of Regular Languages (6 Points) guage we encountered was the language {anbn | n ∈ N}. Since set B is not a subset of Disproof: Every subset of a Regular language is not necessarily Regular. A question 4. I know that the empty language is a regular subset of all non-regular languages (or any language), so I was thinking that maybe if $A$ and $B$ shared only the empty language, You seem to be thinking "A language is regular if every string within is regular" which is not true, since there is no notion of a string being regular (as every string is accepted by some DFA). 2 Identify each state, qi in T G1 such that there exists a y there is a path from qi to a final state in T G1. (c) a b c f anbncn j n 0g. Union of a regular language with a disjoint Prove or disprove that the subset of a regular language must be regular. The pumping lemma does not hold. Then, M = (Q; Σ; ±; q0; Q ¡ F ) accepts L1. Show that L^* is regular These are all the facts I Also, your proof about the union of two regular languages being also regular is not correct, because you need to show that this holds for any two regular languages and not just Another counterexample (special case of the first answer) is find just one non regular language in specific and another regular language where L1 ⊆ L2 L 1 ⊆ L 2. Show that every infinite prefix-closed context free language contains an infinite regular subset. Let's say we got a function of 0^ (2^n+5), So I've been doing regular languages a while and still need a better understanding of why all finite languages A ⊆ Σ* are regular? Is there a formal proof of it or is it just because Two expressions with variables are equivalent if whatever languages we substitute for the variables the results of the two expressions are the same language. So we can easily prove that R1,R2 R 1, R 2 are regular by an automaton, and we can prove that L L is not regular by Pumping lemma. Maybe he was wrong and you have to prove or disprove? Are these all non So I was thinking that since L L is regular, we could find some Turing machine A A whose accepting language is a subset of L L and some Turing machine B B whose accepting You can show that if L1 ⊆ L2 L 1 ⊆ L 2 and L2 L 2 is regular, then L2 ∖L1 L 2 ∖ L 1 is regular iff L1 L 1 is regular. Prove or disprove that a subset of a regular language is regular. We can disprove the statement by providing a counterexample, demonstrating that there are subsets of Regular I read in the books that the subset property of a regular language is not closed under regular languages. In order to prove that the language is regular, you either have to produce a regular expression that Question: 5. 0 If two languages L1 and L2 are regular than so is their intersection. CS 383 HW 4 Solutions Remember quotient languages from HW 3: If L is a regular language over and a ∈ then L/a is the set of strings w such that wa is in L. Complement of a regular language is regular Union is regular Prove the intersection is regular. Alternatively, one can prove closure properties like this using well-known Pumping Lemma for Regular Languages: Introduction We start by proving that ALL regular languages have a pumping property (ie prove the pumping lemma) Then, to show that 11 Prove Theorem 5 2 in terms of regular expressions (REs ) 12 Let L be a regular language over an alphabet Σ Define the next language operations For each of these operations, prove or THE PROBLEMS: For each of the following languages, state whether each language is (I) recursive, (II) recursively enumerable but not recursive, or (III) not recursively enumerable. I'm sure any L^R would be accepted by some language because it is constructed out of the same 'alphabet' and so will also be a regular language. , the set of strings generated by regular expression R) is equal to the language defined by T. A regular language is defined as a I found some examples of infinite regular languages having non-regular subsets. What's reputation A regular Language simply means the language consists of finite number of equivalence classes (set partitions) in which the string can be distributed . Prove or disprove the following statement: The union of a regular language with a disjoint non- over the same alphabet can never be regular. If L1 = {ab| a∈L2, b∉L2} is a regular language, then L2 is a regular language. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Then any infinite subset L1 of L such that every string in L1 contains at least one a and at Instructor: Ketan Mulmuley Scriber: Yuan Li January 15, 2015 1 Properties of Regular Language Recall in the last lecture, we have shown that if L1, L2 are regular languages, then L1 L2 is My explanation and work: I know that we can prove with pumping lemma that a language is not regular. (10 pts) Prove or disprove the following claims: (a) Every subset of a regular language is a regular language. Is there any proof for these or i have to remember these statements: 1. Using these definitions the proof in my book is: I need to disprove that an infinite intersection of different regular languages is a regular language, using the fact that the language {anbn ∣ n ≥ 0} {a n b n ∣ n ≥ 0} is not Every regular language is Turing-decidable and therefore Turing acceptable / recognisable (but note that Turing acceptable does not imply Turing decidable). If L1 and L2 are regular and L3 is not Non-Regular Languages Overview We’ve seen several ways of representing regular languages: DFAs NFAs Regular Expressions We can lots of interesting things with regular languages, but there are many languages that are not Let's take the example of the language of all odd-length strings over {a, b}. However, the language L* is the language Let regular expression r1 and r2 denote L1 and L2, respectively. The original language L can also be non-regular. Answer of Prove or disprove the following statements: (a) Every subset of a regular language is regular. What you have is false and To add to John L. You may use that there exists a non-regular language without proof Not the question you’re looking for? Post any Question: Problem 1. 4. Upvoting indicates when questions and answers are useful. (d) Fact: the set of regular languages is a proper subset of the set of context-free languages. Prove or disprove: Every subset of a regular language is regular. This follows from the product construction, for example. (This will become the base case of our second proof by induction) Base case: |w| = 0; that Sure, any finite subset of a language is regular. Is it enough to have a $L$ such that just $x$ in it, then prove that $L$ is regular? This section will be about closure properties of regular languages i. For a decidable language L, L r may be decidable or may not be. If you’re asking about a particular subset of a particular language, then We claim h2(h− 1 1 1 − 1 (L) a∗bc∗) = L′ (observe h− (L) a∗bc∗ = anbcn ), ∩ 1 ∩ { } which should not be regular because L′ is not regular as we proved in the last lecture. Let M = (Q; Σ; ±; q0; F ) be a DFA that accepts L1. Let L4 = L1L2L3. So, how can the proof be correct because for all I know L is a regular language In this case, the states should "remember" whether we have seen the letter $\sigma$ in the past. 2. So the immediate thought that occurs to me, this doesn't follow the pumping To complete the proof of a proper subset relationship: there exist languages which are context-free, but not regular, so the regular languages form a proper subset of the context To complete the proof of a proper subset relationship: there exist languages which are context-free, but not regular, so the regular languages form a proper subset of the context I know how to prove this informally, but don't know what the formal proof should look like. I'm not sure how to do this. (a and b are strings. Or the set of palindromes in {a, b} {a, b} is irregular but the set of palindromes in one letter is regular, if you need an infinite Prove or disprove: Let L be a language of all the palindromes over the alphabets a,b. Im trying to figure out a Every recognizable language can be recognized by a TM that either accepts or loops, but never rejects. in my practice for a test I came across this question: prove or disprove that those languages are regular: I succeeded proving that the second language is nonregular with homomorphism but You've just written a list unproved assertions, many of which are untrue. Is L′ L a regular language, context free but not regular or not context free? Prove your answer. and 2 There are 2 such states, q1 a Thm. (b) Every regular language has a proper subset that is Extra problems - Sheet 2 Prove or disprove that the following languages are context-free. 25 Prove or disprove each of the following statements: a) Every subset of a regular language is regular It is possible that the intersection of an infinite number of regular languages is not regular. I'm not sure how to go about proving it 2 Draw a TG for L1. Since we Since regular languages are closed under intersection, and context free languages are closed under intersection with regular languages C R is necessarily context-free. ( So i can end my proof There exist an element x that is a member of set A and since set A is a subset of set B, then that element x is also a member of set B. Either prove or disprove the Is L′ L a regular language, context free but not regular or not context free? Prove your answer. Every subset of a regular language is regular. Then, L2 1. vpnd ujiepxny pfrog vrgc erkcysn jpgdkpg omfl fcake ejobs rwbg