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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-12
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Kleene star is the least concatenation-closed language containing L and ε

Statement

Let L be a language over an alphabet Σ.

  1. εL.
  2. LLL.
  3. If K is a language with εK and LKK, then LK.

In particular, L is the least language containing ε and closed under left concatenation by words from L; therefore it is also the least concatenation-closed language containing L and ε.

Facts & Assumptions

Given: A language LΣ and a language KΣ.

[L1]

The powers of L satisfy L0={ε}, for n1 the set Ln consists of words u1un with each uiL, and L=nNLn by Language concatenation, powers, and Kleene star.

[L2]

Language concatenation is AB={ab:aA and bB} by Language concatenation, powers, and Kleene star.

Proof

technique · direct
1.1

By [L1], εL0L. This proves clause 1.

givenL1
1.2

Let wLL. By [L2] there are uL and vL with w=uv. Since vL, [L1] gives some nN with vLn. Then wLn+1 by the definition of Ln+1 in [L1], so wL again. Thus LLL, proving clause 2.

L1L2
1.3

Assume εK and LKK. We prove by induction on nN that LnK. For n=0, [L1] gives L0={ε}K.

givenL1base
1.4

For the induction step, assume LnK. If wLn+1, then by [L1] there are uL and vLn with w=uv. The induction hypothesis gives vK, so wLKK. Hence Ln+1K.

ihL1L2induction
1.5

If u,vL, choose m,nN with uLm and vLn. Writing u=u1um and v=v1vn with every ui,vjL, the concatenation uv is the (m+n)-fold product u1umv1vn, so uvLm+nL. Therefore LLL, and L is concatenation-closed.

L1L2construct
1.6

Any concatenation-closed language K containing L and ε satisfies LKK, because for uLK and vK the concatenation uv lies in KKK.

L2
2.1

By steps 1.3 and 1.4, every Ln is contained in K. Since L=nNLn by [L1], we get LK. This proves clause 3.

step 1.3step 1.4L1discharge-induction
3.1

Step 1.1 gives εL, and [L1] gives L=L1L. Together with step 1.5, this shows that L is itself a concatenation-closed language containing L and ε. Now step 2.1 and step 1.6 show that every other such language contains L. This is exactly the stated minimality.

L1step 1.1step 2.1step 1.5step 1.6

Depends on

Used by

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Dependency tree · two levels

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Sources