LemmaStatement: Literature-sourcedProof: 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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Language concatenation is associative

Statement

For languages $L_{1}, L_{2}, L_{3}$ over the same alphabet, $(L_{1} L_{2}) L_{3} = L_{1} (L_{2} L_{3}) .$

Facts & Assumptions

Given: Languages $L_{1}, L_{2}, L_{3}$ over a fixed alphabet $Σ$ .

[L1]

Language concatenation is $L K = {u v : u \in L and v \in K}$ by Language concatenation, powers, and Kleene star.

[L2]

If $u : m \to Σ$ and $v : n \to Σ$ are words, then the concatenated word $u v : (m + n) \to Σ$ is defined by the offset rule of Computation alphabets, words, the empty word, and $Σ^{*}$ .

Proof

technique · direct

1.1givenL1

Let $w \in (L_{1} L_{2}) L_{3}$ . By [L1] there are words $u \in L_{1}$ , $v \in L_{2}$ and $x \in L_{3}$ with $w = (u v) x$ .

1.2L2algebra

Let $∣ u ∣ = m$ , $∣ v ∣ = n$ , and $∣ x ∣ = p$ . By [L2], both $(u v) x$ and $u (v x)$ have length $m + n + p$ , and for each index $i < m + n + p$ they take the same value: from $u$ when $i < m$ , from $v$ when $m \leq i < m + n$ , and from $x$ when $m + n \leq i < m + n + p$ . Hence $(u v) x = u (v x)$ .

1.3givenL1

Conversely, let $w \in L_{1} (L_{2} L_{3})$ . By [L1] there are words $u \in L_{1}$ , $v \in L_{2}$ , and $x \in L_{3}$ with $w = u (v x)$ .

2.1step 1.1step 1.2L1

Since $v \in L_{2}$ and $x \in L_{3}$ , the word $v x$ lies in $L_{2} L_{3}$ by [L1]. Together with step 1.2 this gives $w = u (v x) \in L_{1} (L_{2} L_{3})$ .

2.2step 1.3L1L2

The same index comparison as in step 1.2 gives $u (v x) = (u v) x$ , and now $u v \in L_{1} L_{2}$ by [L1]. So $w \in (L_{1} L_{2}) L_{3}$ .

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove both inclusions, so $(L_{1} L_{2}) L_{3} = L_{1} (L_{2} L_{3})$ .

Depends on

Used by

A concrete calculation of (L₁L₂)L₃=L₁(L₂L₃) Example

Dependency tree · two levels

6 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

Jean Gallier and Jocelyn Quaintance, Introduction to the Theory of Computation: Some Notes for CIS511 (standard reference, not scraped)