GraphConjecture143.lean`

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3 declarations, 0 contain sorry, 0 declared as axioms

Written on the Wall II - Conjecture 143

Reference: E. DeLaVina, Written on the Wall II, Conjectures of Graffiti.pc

Imports (1)

FormalConjecturesUtil

theorem WrittenOnTheWallII.GraphConjecture143.conjecture143 WOWII Conjecture 143: For a simple connected graph line 55 source
Source documentation
WOWII Conjecture 143:
For a simple connected graph $G$, $\mathrm{tree}(G) \ge (\mathrm{girth}(G) + 1) / \sigma(G)$, where $\mathrm{tree}(G)$ is the largest induced tree size, $\mathrm{girth}(G)$ is the length of the shortest cycle, and $\sigma(G) = G.\mathrm{secondSmallestDegree}$ is the second-smallest degree of $G$'s degree sequence (per WOWII defEntry 65). We state the inequality in denominator-free form to avoid the $\sigma = 0$ corner case ($n \le 1$).
The proof splits into the acyclic case and two cyclic cases. For an acyclic graph the girth is zero. If $\sigma \ge 2$, deleting two consecutive edges from a shortest cycle leaves an induced tree on $\mathrm{girth}(G)-1$ vertices, and the factor $\sigma \ge 2$ yields the required inequality. If $\sigma = 1$, there are two degree-one vertices. A maximal induced tree containing them must have an external vertex with two attachments; those attachments create a cycle whose length forces the tree to contain at least $\mathrm{girth}(G)+1$ vertices.

L4DEL explanation
WOWII Conjecture 143: For a simple connected graph $G$, $\mathrm{tree}(G) \ge (\mathrm{girth}(G) + 1) / \sigma(G)$, where $\mathrm{tree}(G)$ is the largest induced tree size, $\mathrm{girth}(G)$ is the length of the shortest cycle, and $\sigma(G) = G.\mathrm{secondSmallestDegree}$ is the second-smallest degree of $G$'s degree sequence (per WOWII defEntry 65). We state the inequality in denominator-free form to avoid the $\sigma = 0$ corner case ($n \le 1$). The proof splits into the acyclic case and two cyclic cases. For an acyclic graph the girth is zero. If $\sigma \ge 2$, deleting two consecutive edges from a shortest cycle leaves an induced tree on $\mathrm{girth}(G)-1$ vertices, and the factor $\sigma \ge 2$ yields the required inequality. If $\sigma = 1$, there are two degree-one vertices. A maximal induced tree containing them must have an external vertex with two attachments; those attachments create a cycle whose length forces the tree to contain at least $\mathrm{girth}(G)+1$ vertices.
Sources and attribution: Pinned Lean source (1646ca16afd6)

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Fintype, DecidableEq, Nontrivial, SimpleGraph, DecidableRel, SimpleGraph.Adj, SimpleGraph.Connected, LT.lt, Nat, OfNat.ofNat, SimpleGraph.secondSmallestDegree, LE.le, Real, HAdd.hAdd, SimpleGraph.girth, HMul.hMul, SimpleGraph.largestInducedTreeSize
Proof uses: dite, SimpleGraph.IsAcyclic, Classical.propDecidable, SimpleGraph.one_le_largestInducedTreeSize, Eq.mpr, id, Eq, congrArg, SimpleGraph.IsAcyclic.girth_eq_zero, Nat.mul_pos, Not, SimpleGraph.three_le_girth, Exists, And, Ne, SimpleGraph.degree, Membership.mem, Set, SimpleGraph.neighborSet, SimpleGraph.exists_distinct_degree_one_of_secondSmallestDegree_eq_one and 15 more
71 instances, coercions and auxiliary constants left out.

Tactic trace: 20 steps
Open to load the checked before-and-after goal states.
```
theorem conjecture143 (G : SimpleGraph α) [DecidableRel G.Adj] (h : G.Connected)
    (hσ : 0 < secondSmallestDegree G) :
    (G.girth : ℝ) + 1 ≤ (largestInducedTreeSize G : ℝ) * (secondSmallestDegree G : ℝ) := by
…
```
example WrittenOnTheWallII.GraphConjecture143.(example at line 88) line 88 source
Source documentation
The largestInducedTreeSize is nonneg.
```
example (G : SimpleGraph (Fin 3)) : 0 ≤ largestInducedTreeSize G := Nat.zero_le _
…
```
example WrittenOnTheWallII.GraphConjecture143.(example at line 92) line 92 source
Source documentation
The second-smallest degree is nonneg.
```
example (G : SimpleGraph (Fin 3)) [DecidableRel G.Adj] : 0 ≤ secondSmallestDegree G :=
…
```