CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A nonregular language can still satisfy the regular pumping lemma

Statement refuted

Every language satisfying the regular pumping conclusion is regular.

Facts & Assumptions

Given: The language $G : = {w w^{R} v : w, v \in {0, 1}^{+}, w \neq ε} .$

[L1]

By FALSE: the pumping lemma characterizes the regular languages, the claim would be true only if every pumpable language were regular.

[L2]

By Infinitely many pairwise distinguishable prefixes force nonregularity, it is enough to exhibit infinitely many pairwise distinguishable prefixes in $G$ .

[L3]

For $p_{n} = (01)^{n}$ and $x_{n} = (10)^{n} 1$ , the companion refutation proves $p_{n} x_{n} \in G$ and $p_{m} x_{n} \notin G$ whenever $m > n$ : the unique doubled pair in $(01)^{m} (10)^{n} 1$ forces any palindromic prefix $r r^{R}$ to have length $4 m$ , which leaves no room for the required nonempty suffix (FALSE: the pumping lemma characterizes the regular languages).

Counterexample

technique · direct

1.1

The same decomposition used in FALSE: the pumping lemma characterizes the regular languages shows that $G$ satisfies the regular pumping conclusion: words with $∣ w ∣ \geq 2$ pump on the first letter of the front copy of $w$ , while words with $∣ w ∣ = 1$ pump inside the nonempty tail $v$ .

L1given

1.2

For each $n \geq 1$ , the prefix $p_{n} = (01)^{n}$ is distinguished from every longer $p_{m} = (01)^{m}$ by the suffix $x_{n} = (10)^{n} 1$ , because [L3] gives $p_{n} x_{n} \in G$ and $p_{m} x_{n} \notin G$ when $m > n$ . Hence the prefixes $p_{n}$ are pairwise distinguishable, so [L2] shows that $G$ is not regular.

L2L3

2.1

Thus $G$ is a concrete nonregular language that still satisfies the pumping conclusion, so it is a counterexample to the refuted statement.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Eric Blais, Models of Computation, 20. Nonregular Languages (standard reference, not scraped)
Jean Gallier and Jocelyn Quaintance, Introduction to the Theory of Computation: Some Notes for CIS511 (standard reference, not scraped)