SuffixPrefixAvoidance.lean`

Explanation legend Lean4 Domain Explanation Language (L4DEL)

L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.

L4DEL: Lean4 Domain Explanation Language

Annotation provenance. This overlay contains 10 curated AI-written annotations and 7190 mechanical source-derived annotations. They have not been human-reviewed as explanations; treat them as unverified and not mathematical authority. Retained generation history: The 71 original curated annotations were produced in a Claude Sonnet 5.5 session at medium effort; Claude Opus 5.5 is recorded in that session as the advisor model, not the direct author. The bulk source-derived expansion was generated mechanically from pinned source comments by tooling developed in a GPT-5.6 Sol session at ultra reasoning; GPT-5.6 Sol did not originate the quoted source prose. On 2026-10-17 the mechanically generated source-comment annotations of the Formal Conjectures, Carleson, FLT, PFR, CompFiles, Mathematics in Lean and Theorem Proving in Lean overlays were regenerated by the same generator after three corrections made in a Claude Sonnet 5.5 session: titles no longer cut a sentence inside mathematics or end on a dangling word, workflow markers are removed from the quoted prose, and declaration names follow the scanner's namespace resolution. The boilerplate step line was dropped, and comments that were only a row of dashes no longer yield annotations. No new prose was written and no human review is claimed.

Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.

Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

6 declarations, 2 contain sorry, 0 declared as axioms

Suffix-prefix avoidance bound

Let $A$ and $B$ be sets of words of length $n$ over an alphabet with $q$ letters. If no (nonempty) suffix of any word in $A$ coincides with a prefix of any word in $B$, then $$|A| \cdot |B| \leq \frac{q^{2n}}{en}.$$

References:

X post by Dmitry Rybin
[Maximal sets of strings with no prefix-suffix overlap] (https://mathoverflow.net/questions/508648/maximal-sets-of-strings-with-no-prefix-suffix-overlap) by Dmitry Rybin, MathOverflow (2026)
An isoperimetric inequality for word overlap by Dmitrii Zakharov (2026)

Imports (1)

FormalConjecturesUtil

def SuffixPrefixAvoidance.wordSuffix The suffix of a word w : Fin n → Fin q of length k + 1 (the last k + 1 characters) line 45 source
Source documentation
The suffix of a word w : Fin n → Fin q of length k + 1 (the last k + 1 characters).

standard axioms: Quot.sound, propext what this means
Statement uses: Nat, Fin, HAdd.hAdd, OfNat.ofNat
Proof uses: Fin.mk, HSub.hSub
11 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
def wordSuffix (w : Fin n → Fin q) (k : Fin n) : Fin (k + 1) → Fin q :=
…
```
def SuffixPrefixAvoidance.wordPrefix The prefix of a word w : Fin n → Fin q of length k + 1 (the first k + 1 characters) line 49 source
Source documentation
The prefix of a word w : Fin n → Fin q of length k + 1 (the first k + 1 characters).

standard axioms: Quot.sound, propext what this means
Statement uses: Nat, Fin, HAdd.hAdd, OfNat.ofNat
Proof uses: Fin.mk
9 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
def wordPrefix (w : Fin n → Fin q) (k : Fin n) : Fin (k + 1) → Fin q :=
…
```
def SuffixPrefixAvoidance.IsSuffixPrefixAvoiding Two sets of words $A, B$ over Fin q of length n are suffix-prefix avoiding if no nonempty suffix of any word in $A$ equals any prefix of any word in $B$ of the same length line 54 source
Source documentation
Two sets of words $A, B$ over Fin q of length n are suffix-prefix avoiding if no nonempty suffix of any word in $A$ equals any prefix of any word in $B$ of the same length.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Nat, Finset, Fin
Proof uses: Membership.mem, Ne, HAdd.hAdd, OfNat.ofNat, SuffixPrefixAvoidance.wordSuffix, SuffixPrefixAvoidance.wordPrefix
6 instances, coercions and auxiliary constants left out.
```
def IsSuffixPrefixAvoiding (A B : Finset (Fin n → Fin q)) : Prop :=
…
```
theorem SuffixPrefixAvoidance.words_naive_bound $A$ and $B$ are sets of words of length $n$ over alphabet with $q$ letters line 62 source
Source documentation
$A$ and $B$ are sets of words of length $n$ over alphabet with $q$ letters. Trivially then $|A| \cdot |B|$ is at most $q^{2n}$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Nat, Finset, Fin, LE.le, HMul.hMul, Finset.card, HPow.hPow, OfNat.ofNat
Proof uses: of_eq_true, Eq.trans, True, congrArg, HAdd.hAdd, two_mul, pow_add, Fintype, Eq, Fintype.card, eq_true, LE.le.trans_eq, Finset.card_le_univ, congrFun', Finset.prod, Finset.univ, Fintype.card_pi, Finset.prod_congr, Eq.refl, Membership.mem and 3 more
32 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem words_naive_bound
    (A B : Finset (Fin n → Fin q)) :
    A.card * B.card ≤ q ^ (2 * n) := by
…
```
theorem SuffixPrefixAvoidance.suffix_prefix_avoidance_bound $A$ and $B$ are sets of words of length $n$ over alphabet with $q \geq 1$ letters sorry line 76 source
Source documentation
$A$ and $B$ are sets of words of length $n$ over alphabet with $q \geq 1$ letters. No suffix of a word in $A$ coincides with a prefix of a word in $B$. Then $|A| \cdot |B|$ is at most $\frac{q^{2n}}{en}$.
This problem is from Maximal sets of strings with no prefix-suffix overlap and was proved in An isoperimetric inequality for word overlap.

depends on sorry what this means
Statement uses: Nat, Finset, Fin, LT.lt, OfNat.ofNat, SuffixPrefixAvoidance.IsSuffixPrefixAvoiding, LE.le, Real, HMul.hMul, Finset.card, HDiv.hDiv, HPow.hPow, Real.exp
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
34 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem suffix_prefix_avoidance_bound
    (A B : Finset (Fin n → Fin q))
    (hq : 0 < q) (hn : 0 < n)
    (h : IsSuffixPrefixAvoiding A B) :
    (A.card : ℝ) * B.card ≤ (q : ℝ) ^ (2 * n) / (exp 1 * n) := by
…
```
Mentions: SuffixPrefixAvoidance.IsSuffixPrefixAvoiding
theorem SuffixPrefixAvoidance.suffix_prefix_avoidance_weaker_bound sorry line 90 source
depends on sorry what this means
Statement uses: Nat, Finset, Fin, LT.lt, OfNat.ofNat, SuffixPrefixAvoidance.IsSuffixPrefixAvoiding, LE.le, Rat, HMul.hMul, Finset.card, HDiv.hDiv, HPow.hPow
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
24 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.