L = {(01)n(10)m | n ≥ 0, m ≥ 2} is a regular language. Select the errors that are present in the following fooling set ‘proof’ that L is not regular:

Proof (attempt):
Suppose there is a DFA with k states that recognises L. Consider the set S = {(01)n | 1 ≤ n ≤ k}.
Every pair of strings in S can be written as (01)i and (01)j, with i

Question

L = {(01)n(10)m | n ≥ 0, m ≥ 2} is a regular language. Select the errors that are present in the following fooling set ‘proof’ that L is not regular:

Proof (attempt):
Suppose there is a DFA with k states that recognises L. Consider the set S = {(01)n | 1 ≤ n ≤ k}.
Every pair of strings in S can be written as (01)i and (01)j, with i<j. These two strings are distinguishable with respect to L with the distinguishing suffix (10)i, as (01)i(10)i is in L but (01)j(10)i isn't.

Thus every pair of distinct strings in S is distinguishable with respect to L, and so S is a fooling set of size k. Thus, there cannot exist a DFA with k states that recognises L. Because k was arbitrary, no DFA that recognises L can exist, and so L is not regular.

BlackTom AI · Accepted Answer

Question restatement: We are given the language L = {(01)^n(10)^m | n ≥ 0, m ≥ 2} and a proposed fooling-set-style argument to claim L is not regular. The fooling set S is defined as S = {(01)^n | 1 ≤ n ≤ k}, with the claim that for i < j, the strings (01)^i and (01)^j are distinguishable by the suffix (10)^i, which would imply S is a fooling set of size k and hence L is not regular. We are asked to identify errors present in this proof attempt from the provided answer options.

Option 1: It incorrectly states that a string is not a member of L.
- Analysis: The core issue in the fooling-set argument is not a stated claim that a particular string is or isn’t in L; rather, it incorrectly uses the set S as though its elements are in L, and it relies on a suffix (10)^i to distinguish between (01)^i and (01)^j. The strings in S are of the form (01)^n with no trailing (10)^m, which do not satisfy the form (01)^n(10)^m with m ≥ 2. Therefore, these S-elements are not in L. However, the statement in this option says the fooling-set argument “incorrectly states that a string is not a member of L.” The issue is more precise: the fooling set is constructed using strings that are not in L, which means the basis of the fooling-set argument is flawed because a fooling set for a language L typically consists of strings that are in L and that can be extended to stay outside L in a controllable way. Saying that the proof “specifically states” that a string is not in L is not the standard way the error is framed; the deeper error is that S itself is not a subset of L. Therefore, this option is somewhat misleading or misframed as a standalone error description. Verdict: not a precise or reliable identification of the error; arguably not the clearest standalone incorrect step. Consider this option as not a solid standalone diagnostic.

Option 2: The suffix given for each distinct pair in S isn't always distinguishing.
- Analysis: This is a precise and real problem. The distinguishing suffix proposed is (10)^i applied to the two strings (01)^i and (01)^j with i < j. For i = 1, (01)^1(10)^1 equals 01 10, which is not in L since the required m is at least 2. Therefore, the suffix (10)^i does not distinguish the pair ((01)^i, (01)^j) when i = 1, because the resulting string (01)^i(10)^i is not in L regardless of j. This shows the claim that all pairs are distinguishable using that suffix is false. Verdict: Correct identification of an error.

Option 3: It does not give a concrete number for k.
- Analysis: The critique here would be that the argument relies on k as an arbitrary parameter and then concludes non-regularity by letting k be arbitrary. In standard fooling-set proofs, you typically need a fooling-set of size greater than the number of states to derive a contradiction for any DFA; merely having a parameter k and not fixing a specific numeric bound is not itself a mathematical flaw if the construction were valid for all k. The absence of a concrete number for k is not per se an error in the logical structure of a fooling-set argument; it is common to argue for all k or to derive a contradiction in terms of an unbounded family. Verdict: Not a solid standalone error; the lack of a concrete numeric value does not by itself invalidate the reasoning, given the quantification over k is part of the standard argument. Therefore this option is not an essential error.

Option 4: The size of the given fooling set is not greater than k.
- Analysis: This is a key point. The set S has size k (since n ranges from 1 to k). In fooling-set arguments intended to prove non-regularity, one typically requires a fooling set of size greater than the number of states of any DFA (i.e., size > k for a k-state DFA). Here, the claimed fooling set has size exactly k, which does not suffice to force a contradiction for a k-state DFA. Hence the conclusion that no DFA with k states recognises L is invalid; to prove non-regularity you would need a fooling set of size k+1 or larger. Verdict: Correct identification of an error.

Option 5: The fooling set contains strings not in L.
- Analysis: This is a fundamental structural flaw. The fooling-set construction presumes that the elements of S are within the language L, so that the distinguishing suffixes behave in the intended way relative to membership in L. However, each string in S is of the form (01)^n with no trailing (10)^m, which does not conform to the language L’s form (01)^n(10)^m with m ≥ 2. Therefore, S ∉ L, meaning the proposed fooling set is not a valid fooling set for L. This is a core error in the reasoning. Verdict: Correct identification of an error.

Summary of evaluations:
- Option 1: Not a precise standalone diagnostic; not the clearest description of the error. (Not a solid standalone error claim.)
- Option 2: Correct; the proposed distinguishing suffix does not work for i = 1, so not all pairs are distinguished as claimed.
- Option 3: Not an essential or definitive error in the standard fooling-set framework; not a firm flaw by itself.
- Option 4: Correct; the fooling set size being equal to k (not greater than k) does not suffice to force a contradiction for a k-state DFA.
- Option 5: Correct; the fooling set contains strings not in L, which invalidates the fooling-set argument.

Putting it together, the most robustly correct identifications of errors are options 2, 4, and 5. Option 1 is somewhat misframed as a statement about what is “stated,” and option 3 does not constitute a decisive mathematical flaw in the fooling-set approach.

COMP30026_2025_SM2 Worksheet 8

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