Let Σ = {0, 1}. Show that: (a) the language L₂ = {0k s0k|k ≥ 1 and s € Σ*} is regular. (b) the language L3 = {0k1s0k|k ≥ 1 and s € Σ*} is not regular.

Q: Prove that the following language A is decidable (e.g., give a TM that decides A). A = {⟨M⟩ | M is…

A: We will prove that language A = {⟨M⟩ | M is an NFA and L(M) = B}, where B is a regular language, is…

Q: Build a deterministic FA M2 for the following language L2=L(M2)= {x over {a, b, c} | x has at least…

A: This DFA can be made using for States. Below is the answer

Q: a. L= {x€{a, b, c }* | na(x) < n»(x) and na(x) < nc(x) }, where n(x) is the number of occurrences of…

A:

Q: Let L = {w#w | w ∈ {a, b}*} be a language. Use the pumping lemma for regular languages to prove that…

A: L = {w#w| w∈ {a, b}*} string generated by this languageare aaab#aaab, babb#babb proof: If possible…

Q: Define a CFG that generates the language over the alphabet Σ = {α,β,θ} defined by the regular…

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Q: (3) {w € [0, 1]*: w has 001 as a substring} Testing strings: €, 0, 1, 01, 000, 001, 0001, 0010,…

A: We are given a language {w ∈ {0,1}*: w has 001 as a substring}.And we are asked to test given…

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Q: Prove that the following language is undecidable by giving a reduction from a known undecidable…

A: SOLUTION: The language to show undecidable is  EQCFG=G,HL(G) = L(H). Consider the following…

Q: Given the grammar    F>   GnF | mZpG G    >    pZ | mZpM | q    Z    >    mG | mGp | m…

A: Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: Let L be a regular language. Show that the following language L' is regular too. L' = {y : 3x € £*…

A: | x | = | y | = | z | xyz € L  L is regular then L' is regular.

Q: Consider the following languages over the alphabet Σ = {a, b}: 1. L1 = {xwx : x, w ∈ Σ∗} 2. L2 =…

A: Making the distinction between regular and non-regular languages is a crucial task in the field of…

Q: Consider the following languages over Σ = {0,1} L6 is the language of the NFA given by the…

A: The NFA (Non-Deterministic Finite Automaton) that you have provided defines the language L6. We need…

Q: Question 3. Consider the language L = {w€ {0, 1}* | the strings 01 and 10 occur the same number of…

A: Given language is, L={w ∈{0,1}* | the strings 01 and 10 occur same number of times in w} Pumping…

Q: Let Σ = (a, b, c). Prove that the following languages over alphabet Σ are not regular.

A:

Q: Consider the language L = AEqB = {x ∈ {a, b}∗ | na(x) = nb(x)}. a. Show that if na(x) − nb(x) =…

A: Answer: I have given answered in the handwritten format in brief explanation

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Q: Think about the concept of extended transition functions δ* and ∆* of Deterministic Finite Automata…

A: Think about the concept of extended transition functions δ* and ∆* of Deterministic Finite Automata…

Q: Give a regular expression for the smallest language L over Σ = {0, 1} such that •€¢ L 0 EL ● ● 1 EL

A: Since you have posted a question with multiple sub parts, we will provide the solution only to the…

Q: Prove that the following language is undecidable by giving a reduction from a known undecidable…

A: Actually, the given information is: ????? = {⟨?, ?⟩ | ? ??? ? ??? ???? ???? ???? ?(?) = ?(?)}Hint:…

Q: U̲s̲i̲n̲g̲ ̲t̲h̲e̲ ̲p̲u̲m̲p̲i̲n̲g̲ ̲l̲e̲m̲m̲a̲,̲ ̲s̲h̲o̲w̲ ̲t̲h̲a̲t̲ ̲t̲h̲e̲ ̲f̲o̲l̲l̲o̲w̲i̲n̲g̲…

A: Pumping Lemma states that if the infinite language has to be accepted by a finite automata, then…

Q: (4) {w € [0, 1]* : w does not have 001 as a substring} Testing strings: €, 0, 1, 01, 000, 001, 0001,…

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Q: Show that the language is not context-free. L = {an-1bncn+1: n ≥ 1}

A: Below I have provided the handwritten solution to the given question

Q: Suppose I is a non empty regular language, over Σ, such that every w E L satisfies [w] = k, for some…

A: Given that, non-empty regular language L over the alphabet Σ such that every string w in L satisfies…

Q: Consider the language L = {a'b'c* | i # j,j # k, i # k}; it is not a context- free language. Show…

A: Pumping lemma for CFL is used to prove the language is not cfl using some valid string s.  Never use…

Q: Let Z={a,b,c}. Show that (b) The set of all languages over Σ is uncountable

A: We need to prove the set of all languages over sigma is uncountable.

Q: Use the pumping lemma to prove that the language L over {a,b,c}", given by L = {wcv: |wa = |vl6} is…

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Q: Let FH be the set of first halves of the strings in A. FH = {x | xy ∈ A} for some y, and |x| = |y|.…

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Q: Let Σ = {0,1}. Show that {0ku0k | k >= 1 and u ε Σ*} is a regular language

A: Let Σ = {0,1}. Show that {0ku0k | k &gt;= 1 and u ε Σ*} is a regular language

Q: Let L be the language L={ 0n1m | n≠(m-7) } over the alphabet Σ={0,1}. Prove that L is not regular…

A: Let L be the language L={ 0n1m | n≠(m-7) } over the alphabet Σ={0,1}. Prove that L is not regular…

Q: L* is regular, then L must be regular. For any language L, L* must be regular If L1L2 is regular,…

A: Note As Per over guidelines we will solve three sub part remaining sub part post separately i did 2…

Q: If A is any language, let A 1/3 – 1/3 be the set of all strings in A with their middle thirds…

A: the answer is given below:-

Q: Consider the language L = {a^b³ct | r, s, t € N and t ≤r× s} Problem 3 Use the pumping lemma to…

A: Pumping lemma for CFL is used to prove the language is not cfl using some valid string s.  Never use…

Q: Show that the language L = {w E {a,6, c}" : na (w) + nb w) = nc (W)} is

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Q: Consider the following languages over Σ = {0,1} for L7 is the language described by 1∗(011∗)∗…

A: The breakdown of the regular expression is provided below.

Q: Prove that the following language is non-regular, using the Pumping Lemma. L = {amb¹c² | m, n, r =…

A: Regular languages can be recognized by finite automata or described using regular expressions. To…

Q: Suppose L = {wlw does not contain the substring 11 is a language over the alphabet {0, 1}. Prove…

A: To formally prove that the language L = {w | w in {0,1}* and w does not contain the substring "11"}…

Q: This is the Java programming language.  Please respond as soon as possible.  To represent set…

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Q: Consider the language L = { 3x such that x = L(M₁) L(M₂)}. Show that L is not decidable.

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A: