Lean Proof Explanation: IsCycle.three_le

could you explain the lean proof of this theorem

The following references were attached as context:
File path: Mathlib/Combinatorics/SimpleGraph/Path.lean
Source URL: GitHub Link
Repo: leanprover-community/mathlib4
CommitOID: 1a9b964b92509b9a757fd24bae6b6b9812443163

lemma IsCycle.three_le_length {v : V} {p : G.Walk v v} (hp : p.IsCycle) : 3 ≤ p.length := by

Absolutely! The lemma you are referring to is:

lemma IsCycle.three_le_length {v : V} {p : G.Walk v v} (hp : p.IsCycle) : 3 ≤ p.length := by

What does the lemma state?

Context:
- G is a simple graph.
- V is the type of vertices.
- G.Walk v v is a walk in G that starts and ends at the same vertex v.
- p : G.Walk v v is such a walk.
- hp : p.IsCycle means that the walk p is a cycle (it is closed, nontrivial, and has no repeated vertices except for the start/end).
Claim: If p is a cycle, then its length is at least 3.

Why is this true?

A cycle in a simple graph must visit at least 3 vertices (otherwise, it would either be a loop or a repeated edge, which are not allowed in a simple graph).
The minimal cycle is a triangle (3 edges, 3 vertices).

What does a Lean proof look like?

While I can't see the exact proof in the file, a typical Lean approach would be:

Unpack IsCycle: Use the definition of IsCycle to get that p is a nontrivial closed walk with no repeated vertices except at the endpoints.
No Repeated Vertices: The only way to have no repeats except start/end is if the walk visits at least 3 distinct vertices.
Length Calculation:
- The length of a walk is the number of edges.
- For a cycle, this is at least 3 (since fewer would mean repeats or triviality).

A sketch of the proof in Lean:

lemma IsCycle.three_le_length {v : V} {p : G.Walk v v} (hp : p.IsCycle) : 3 ≤ p.length :=
begin
  -- Expand what it means for p to be a cycle
  rcases hp with ⟨hne, h_closed, h_norep⟩,
  -- hne: p is nontrivial (not nil or single edge)
  -- h_closed: p is closed (starts and ends at v)
  -- h_norep: no repeated vertices except endpoints
  -- Use the facts above to show p.length ≥ 3
  -- (Details would involve showing that the minimal nontrivial closed walk with no repeated vertices must have at least 3 edges)
end

Intuitive Explanation

If a cycle had length 1, it would be a loop (not allowed in simple graphs).
If it had length 2, it would go from v to w and back, so vertices would repeat (contradicts "no repeats except endpoints").
So, the minimal possible length for a cycle is 3.

Summary

The lemma states that every cycle in a simple graph has at least length 3.
The proof uses the definition of a cycle (no repeated vertices except at endpoints, closed, nontrivial) to rule out cycles of length 1 or 2.

If you want to see the full Lean code proof from the file, let me know and I can fetch and explain it line by line!

Do please fetch the full Lean code proof and explain it line by line

Here is the full Lean proof for the lemma you asked about, followed by a detailed line-by-line explanation:

lemma IsCycle.three_le_length {v : V} {p : G.Walk v v} (hp : p.IsCycle) : 3 ≤ p.length := by
  have ⟨⟨hp, hp'⟩, _⟩ := hp
  match p with
  | .nil => simp at hp'
  | .cons h .nil => simp at h
  | .cons _ (.cons _ .nil) => simp at hp
  | .cons _ (.cons _ (.cons _ _)) => simp_rw [SimpleGraph.Walk.length_cons]; omega

Let's break this down:

1. Statement and Parameter Unpacking

lemma IsCycle.three_le_length {v : V} {p : G.Walk v v} (hp : p.IsCycle) : 3 ≤ p.length := by

Meaning: If p is a cycle (i.e., p.IsCycle), then the length of p is at least 3.

2. Extracting Structure from the Hypothesis

have ⟨⟨hp, hp'⟩, _⟩ := hp

Here, hp is unpacked using Lean's structure projection and pattern matching.
IsCycle extends IsCircuit, which in turn extends IsTrail, so hp contains:

The trail property (hp),
The non-nil (nonempty) property (hp'),
The property that the only repeated vertex is the start/end.

3. Case Analysis on the Walk Structure

match p with
| .nil => simp at hp'
| .cons h .nil => simp at h
| .cons _ (.cons _ .nil) => simp at hp
| .cons _ (.cons _ (.cons _ _)) => simp_rw [SimpleGraph.Walk.length_cons]; omega

The proof proceeds by splitting into cases depending on the structure of the walk p:
- In Lean, .nil is the empty walk, .cons h p is a step (edge) followed by a subwalk.
Case 1: p is .nil (empty walk)
| .nil => simp at hp'
If p is empty, hp' (which is the non-nil property) is contradicted by simp, so this case cannot actually happen.
Case 2: p is a single edge: .cons h .nil
| .cons h .nil => simp at h
If p is just one edge, then h is an adjacency. But for a cycle, this is impossible in a simple graph (would require a loop), so simp at h derives a contradiction.
Case 3: p is two edges: .cons _ (.cons _ .nil)
| .cons _ (.cons _ .nil) => simp at hp
If p is two edges, this would mean a walk from v to some w and back to v. But for a cycle in a simple graph, this would require a repeated vertex or edge, which is not allowed. So simp at hp finds the contradiction.
Case 4: p is three or more edges: .cons _ (.cons _ (.cons _ _))
| .cons _ (.cons _ (.cons _ _)) => simp_rw [SimpleGraph.Walk.length_cons]; omega
This matches the case where p has at least three edges. simp_rw [SimpleGraph.Walk.length_cons] rewrites the length of the walk in terms of cons cells. omega is an automation tactic that solves linear arithmetic goals, so it checks that the length is at least 3.

Summary

The proof eliminates the cases where the walk is of length 0, 1, or 2, showing that a cycle must have at least length 3.
For length ≥ 3, it is automatically true by the definition of length.

Why is this correct?

In a simple graph, a cycle cannot have fewer than three edges because that would mean a loop or a repeat, both forbidden by the definition of a cycle in this context.

If you’d like clarification on any specific line, or want to know how the supporting definitions (like IsCycle) work, let me know!

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  <title>Lean Proof Explanation: IsCycle.three_le_length</title>
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  <h1>Lean Proof Explanation: <code>IsCycle.three_le_length</code></h1>
  <div class="block user">
    <span class="header">User:</span>
    <p>could you explain the lean proof of this theorem</p>
    <pre>
lemma IsCycle.three_le_length {v : V} {p : G.Walk v v} (hp : p.IsCycle) : 3 ≤ p.length := by
    </pre>
  </div>...

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