Alphabeta Math
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 21 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 21 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Graphs, Walks and Connectivity

1 · Prerequisites

2 · Summary

Finite cardinality supplies graph order, size, degree counts and the enumeration of two-element subsets, while Double counting: ∑x∈X∣Rx∣=∣R∣=∑y∈Y∣Ry∣ for a relation between finite sets supplies the incidence identity behind the handshake lemma. The general definition Equivalence relation, equivalence class, and the quotient set A/∼, from the required Relations, Functions, and Quotients development, is applied to reachability. The counts A finite set with n elements has exactly (n2) two-element subsets, and 2(n2)=n(n−1) and ∣P(A)∣=2∣A∣ for finite A license the complete-graph and labelled-graph enumerations.

A graph is fixed as finite, simple and undirected before adjacency, subgraphs, walks, paths, cycles and connected components are defined. Removing repetitions turns walks into paths; reachability then partitions the vertices, shortest-path distance satisfies the metric axioms componentwise, and parity characterizes bipartite graphs. Standard graph constructions lead to the handshake lemma and its parity consequence. Vertex and edge cuts yield Whitney's inequalities, a bridge is defined as an edge whose deletion adds a component and characterised as an edge on no cycle, and longest-path and extremal edge arguments give cycle and connectivity criteria.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

A finite simple graph is a finite vertex set together with a set of two-element vertex subsets

Definition

A finite simple graph is an ordered pair G=(V,E) in which V is a finite set and

E⊆[V]2:={ {u,v}⊆V:u≠v }.

The elements of V=V(G) are the vertices and the elements of E=E(G) are the edges. Thus every edge has two distinct endpoints, there are no loops, and an unordered pair of vertices occurs as at most one edge. Finiteness and the notation ∣V∣ and ∣E∣ are those of The cardinality ∣A∣ of a finite set.

When no qualifier is present, the word graph has this meaning.

RemarkRemark: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Unless stated otherwise, graph means finite, simple and undirected; orders, sizes and empty-set conventions are fixed here

For a graph G=(V,E) (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets), its order is ∣V∣ and its size is ∣E∣ (The cardinality ∣A∣ of a finite set). The null graph has V=∅ and E=∅. An edgeless graph has E=∅ but may have vertices. Thus the null graph is the unique graph with no vertices, while an edgeless graph need not be null.

All sums indexed by V or E use their ordinary empty values. A minimum or maximum taken over the vertex set is used only when V≠∅; in particular, minimum and maximum degree are not assigned values for the null graph. Connectivity conventions for the null graph and the one-vertex graph are stated with the definition of connectivity.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Multigraphs, loops and directed graphs as variants distinct from the default finite simple graph

Definition

A finite undirected multigraph consists of a finite vertex set V, a finite edge set M, and an endpoint map assigning to each e∈M either a singleton {v} or a two-element subset {u,v} of V. An edge assigned {v} is a loop at v. Distinct members of M assigned the same endpoint set are parallel edges.

A finite directed graph, or digraph, consists of a finite vertex set V and a set A⊆V×V of arcs. An arc (u,v) is directed from u to v; an arc (v,v) is a directed loop.

These objects are variants, not instances of the default graph convention in A finite simple graph is a finite vertex set together with a set of two-element vertex subsets and Unless stated otherwise, graph means finite, simple and undirected; orders, sizes and empty-set conventions are fixed here. A theorem stated merely for a graph therefore neither permits loops or parallel edges nor supplies directions to its edges.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree

Definition

Let G=(V,E) be a graph (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets).

Distinct vertices u,v are adjacent, or neighbours, when {u,v}∈E. A vertex v and an edge e are incident when v∈e. The open neighbourhood and closed neighbourhood of v are

NG(v):={ u∈V:{u,v}∈E },NG[v]:=NG(v)∪{v}.

The degree of v is deg⁡G(v):=∣NG(v)∣, equivalently the number of edges incident with v. A graph is r-regular when every vertex has degree r; it is cubic when it is 3-regular. The multiset of the vertex degrees, usually written in nonincreasing order, is the degree sequence.

When V≠∅, the minimum degree and maximum degree are

δ(G):=min⁡v∈Vdeg⁡G(v),Δ(G):=max⁡v∈Vdeg⁡G(v).

They are defined because V is a nonempty finite set (The cardinality ∣A∣ of a finite set). Neither δ(G) nor Δ(G) is defined for the null graph.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Subgraphs, induced subgraphs and spanning subgraphs

Definition

Let G=(V,E) be a graph. A graph H=(W,F) is a subgraph of G when W⊆V and F⊆E∩[W]2.

For W⊆V, the subgraph induced by W is

G[W]:=(W, E∩[W]2).

Thus it retains every edge of G whose two endpoints lie in W. A subgraph H=(W,F) is induced when H=G[W], and it is spanning when W=V. These constructions remain finite simple graphs by A finite simple graph is a finite vertex set together with a set of two-element vertex subsets.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges

Definition

Let G=(V,E) be a graph. A walk of length ℓ is a finite vertex list

W=(v0,v1,…,vℓ)

such that {vi−1,vi}∈E for every 1≤i≤ℓ. Its endpoints are v0 and vℓ. The length is the number ℓ of traversed edges, not the number of displayed vertices. A walk is closed when v0=vℓ.

A trail is a walk in which the traversed edges are distinct. A path is a walk in which the vertices are distinct. In particular, (v) is a path of length 0 from v to itself.

A cycle is a closed walk (v0,…,vℓ−1,vℓ=v0) of length ℓ≥3 in which v0,…,vℓ−1 are distinct. A graph is acyclic when it contains no cycle. Reversing any walk, trail, path, or cycle preserves its type and length.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31Open item page →

Every walk between two vertices contains a path between the same endpoints

Statement

Let W=(v0,…,vℓ) be a walk in a graph. By deleting zero or more closed segments from its vertex list, one obtains a path from v0 to vℓ of length at most ℓ. If W repeats a vertex, the resulting path can be chosen to have length strictly less than ℓ.

Facts & Assumptions

Given: A walk W=(v0,…,vℓ) in a finite simple graph.

[F2]

If vi=vj with i<j, deleting vi+1,…,vj leaves the list (v0,…,vi,vj+1,…,vℓ), which is a walk with the same endpoints and length ℓ−(j−i)<ℓ.

[F3]

A finite vertex list has only finitely many lists obtainable by deleting segments, so any nonempty family of their lengths has a least member (The cardinality ∣A∣ of a finite set).

Proof

technique · constructive
1.1

Let W be the nonempty finite family of walks obtained from W by deleting zero or more closed segments, and choose P∈W of least length.

givenF2F3chooseconstruct
2.1

The walk P has no repeated vertex: if two positions carried the same vertex, [F2] would delete the intervening closed segment and produce a shorter member of W, contrary to the choice of P.

step 1.1F2
3.1

By [F1], P is a path with the same endpoints as W, and its length is at most ℓ because W∈W. If W repeats a vertex, applying [F2] once before minimizing gives a member of W shorter than W, so P is strictly shorter than W.

step 1.1step 2.1F1F2discharge-construct∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Connected graphs and connected components defined by the existence of vertex paths

Definition

Vertices u and v of a graph G are connected, or reachable from one another, when a path in G has endpoints u and v. Equivalently, a walk joins them, because Every walk between two vertices contains a path between the same endpoints turns such a walk into a path.

For v∈V(G), its connected component is the induced subgraph on

CG(v):={ u∈V(G):u is reachable from v }.

A graph is connected when its vertex set is nonempty and every two vertices are reachable from one another. Thus the null graph is not connected. The one-vertex graph is connected, since its vertex is joined to itself by the path of length 0 from Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Being joined by a walk or path is the same equivalence relation on the vertex set

Statement

On the vertex set of a graph, define u∼v when a walk joins u to v. Then ∼ is an equivalence relation (Equivalence relation, equivalence class, and the quotient set A/∼), and u∼v holds exactly when a path joins u to v.

Facts & Assumptions

Given: A graph G=(V,E) and the relation u∼v defined by the existence of a walk from u to v.

[F1]

An equivalence relation is reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set A/∼).

[L1]

Every walk between two vertices contains a path between the same endpoints (Every walk between two vertices contains a path between the same endpoints).

[F2]

A one-vertex list is a path of length 0, reversing a walk gives a walk with reversed endpoints, and concatenating a u to v walk with a v to w walk gives a u to w walk (Connected graphs and connected components defined by the existence of vertex paths, Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

Proof

technique · direct
1.1

For every u∈V, the length-0 walk (u) joins u to itself, so u∼u.

F2
1.2

If u∼v, reversing a witnessing walk gives a walk from v to u, so v∼u.

givenF2
1.3

If u∼v and v∼w, concatenating witnessing walks gives a walk from u to w, so u∼w.

givenF2
2.1

Steps 1.1, 1.2 and 1.3 prove reflexivity, symmetry and transitivity, hence ∼ is an equivalence relation by [F1].

step 1.1step 1.2step 1.3F1
3.1

Every path is a walk, while [L1] turns every witnessing walk into a path with the same endpoints; therefore walk-reachability and path-reachability coincide.

L1F2∎
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

The connected components of a graph partition its vertex set and are its maximal connected subgraphs

Statement

The vertex sets CG(v) of the connected components of a graph G are nonempty, cover V(G), and any two are equal or disjoint. Each induced graph G[CG(v)] is connected and contains every connected subgraph of G that contains v. Consequently the connected components are precisely the maximal connected subgraphs of G.

Facts & Assumptions

Given: A graph G=(V,E) and the reachability classes CG(v).

[F1]

CG(v) is the set of vertices reachable from v, and the connected component at v is the induced graph G[CG(v)] (Connected graphs and connected components defined by the existence of vertex paths, Subgraphs, induced subgraphs and spanning subgraphs).

Proof

technique · direct
1.1

Reflexivity in [L1] gives v∈CG(v), so every class is nonempty and the classes cover V.

L1F1
1.2

If CG(u) and CG(v) meet at w, then u reaches w and w reaches v by symmetry; transitivity gives that u reaches v, and another use of symmetry and transitivity shows a vertex reaches u exactly when it reaches v. Thus CG(u)=CG(v), so distinct classes are disjoint.

L1F1
1.3

If x,y∈CG(v), then x reaches v and v reaches y; concatenating these paths gives a walk from x to y, and deleting any repeated segments gives a path all of whose vertices remain reachable from v. Hence G[CG(v)] is connected.

L1F1
1.4

If a connected subgraph H contains v, every vertex x of H is joined to v by a path in H, hence also in G, so x∈CG(v). Thus H is a subgraph of G[CG(v)].

F1
2.1

Steps 1.1 and 1.2 give the partition assertion, and steps 1.3 and 1.4 say exactly that its induced subgraphs are connected and maximal among connected subgraphs.

step 1.1step 1.2step 1.3step 1.4∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Graph distance within a component, eccentricity, diameter and girth, including the acyclic convention

Definition

Let u and v lie in the same connected component of a graph G. Their distance is

dG(u,v):=min⁡{ ℓ∈N:some path from u to v has length ℓ }.

The set being minimized is nonempty by connectivity and finite because a path has no repeated vertex and V(G) is finite (The cardinality ∣A∣ of a finite set). No ordinary natural-number distance is assigned to vertices in different components.

For a vertex v, its eccentricity is the maximum of dG(v,u) over the vertices u in its component. The diameter of a connected graph is the maximum of dG(u,v) over all pairs of its vertices. Thus every component has its own diameter, while a disconnected graph has no single graph diameter under this convention.

The girth g(G) is the minimum length of a cycle in G. If G is acyclic, set g(G)=∞ by convention. This symbol records absence of a cycle and is not a distance between vertices.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Shortest-path distance is a metric on every connected component

Statement

Let C be the vertex set of a connected component of a graph G. The function dG:C×C→N satisfies, for all u,v,w∈C,

  1. dG(u,v)≥0, with dG(u,v)=0 if and only if u=v;
  2. dG(u,v)=dG(v,u);
  3. dG(u,w)≤dG(u,v)+dG(v,w).

These are the metric axioms, so shortest-path distance is a metric on C.

Facts & Assumptions

Given: Vertices u,v,w in one connected component C of G.

[F1]

dG(x,y) is the least length of an x to y path, and the length is a natural number (Graph distance within a component, eccentricity, diameter and girth, including the acyclic convention).

[L1]

Every walk contains a path between the same endpoints whose length is no greater (Every walk between two vertices contains a path between the same endpoints).

[F2]

Reversing a path preserves its length, and concatenating an x to y path of length a with a y to z path of length b gives an x to z walk of length a+b.

Proof

technique · direct
1.1

Every path length is a natural number, hence dG(u,v)≥0. The length-0 path (u) gives dG(u,u)=0. Conversely a path of length 0 has one vertex, so dG(u,v)=0 forces u=v.

F1
1.2

Reversing a shortest u to v path gives a v to u path of the same length, so dG(v,u)≤dG(u,v); reversing the roles gives the opposite inequality, hence equality.

F1F2
1.3

Concatenate shortest paths from u to v and from v to w. This is a u to w walk of length dG(u,v)+dG(v,w), and [L1] supplies a u to w path of no greater length. Minimality in [F1] gives dG(u,w)≤dG(u,v)+dG(v,w).

F1F2L1
2.1

Steps 1.1, 1.2 and 1.3 establish separation, symmetry and the triangle inequality directly on C, so dG is a metric there.

step 1.1step 1.2step 1.3∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

A bipartite graph and a proper two-colouring of its vertices

Definition

A graph G=(V,E) is bipartite when there are disjoint subsets A,B⊆V with V=A∪B such that every edge has one endpoint in A and the other in B. The ordered pair (A,B) is a bipartition.

Equivalently, a proper two-colouring is a map c:V→{0,1} satisfying c(u)≠c(v) whenever {u,v}∈E: take A=c−1[{0}] and B=c−1[{1}], or define c from a bipartition. Either part may be empty, so every edgeless graph, including the null graph, is bipartite.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-31Open item page →

Every closed walk of odd length contains a cycle of odd length

Statement

Every closed walk of odd length in a graph contains, among its traversed edges, a cycle of odd length.

Facts & Assumptions

Given: A finite simple graph and its closed walks.

[F1]

A simple graph has no loop, so an odd closed walk has length at least 3; a closed walk of length m≥3 whose vertices before the repeated endpoint are distinct is a cycle (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

[L1]

Mathematical induction: a property true at 0 and inherited from m to m+1 holds for every natural m (The principle of mathematical induction).

[F2]

If a closed walk (v0,…,vr=v0) has vi=vj with 0≤i<j<r, it splits into closed walks of lengths j−i and r−(j−i); when r is odd, exactly one of these lengths is odd.

Proof

technique · induction
1.1

Let P(m) say that every odd closed walk of length at most m contains an odd cycle. At m=0 there is no odd positive length at most 0, so P(0) holds vacuously.

base
1.2

Fix m and assume P(m). Let W=(v0,…,vr=v0) be an odd closed walk with r≤m+1.

ihgiven
2.1

If r≤m, the inductive hypothesis supplies an odd cycle in W.

step 1.2
2.2

Suppose instead that r=m+1. If v0,…,vr−1 are distinct, then W is itself an odd cycle by [F1].

step 1.2F1
2.3

If those vertices are not distinct, choose 0≤i<j<r with vi=vj. By [F2], W splits into two shorter closed walks, and exactly one has odd length. Its length is at most r−1=m, so P(m) supplies an odd cycle in that subwalk and hence in W.

step 1.2F2choose
3.1

Steps 2.1, 2.2 and 2.3 cover every odd closed walk of length at most m+1, so P(m+1) holds.

step 2.1step 2.2step 2.3
4.1

By [L1], P(m) holds for every natural m. Applying it at the length of any given odd closed walk yields an odd cycle contained in that walk.

step 1.1step 3.1L1discharge-induction∎
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31Open item page →

A finite graph is bipartite if and only if it has no odd cycle

Statement

A finite simple graph is bipartite if and only if it contains no cycle of odd length.

Facts & Assumptions

Given: A finite simple graph G=(V,E).

[F1]

A bipartition is an ordered pair (A,B) of disjoint subsets with V=A∪B and every edge running between its two parts (A bipartite graph and a proper two-colouring of its vertices).

[L1]

Every odd closed walk contains an odd cycle (Every closed walk of odd length contains a cycle of odd length).

[F2]

Path-reachability is an equivalence relation, and its equivalence classes are the connected components; hence two vertices lie in the same component exactly when a path joins them. Every such path has a length and therefore a parity (Being joined by a walk or path is the same equivalence relation on the vertex set, Connected graphs and connected components defined by the existence of vertex paths, Graph distance within a component, eccentricity, diameter and girth, including the acyclic convention).

[L2]

The connected components are nonempty, cover the vertex set, and are pairwise equal or disjoint (The connected components of a graph partition its vertex set and are its maximal connected subgraphs).

Proof

technique · direct
1.1

Suppose G has a bipartition (A,B). Along any walk the successive vertices alternate between A and B, so a closed walk returns to its initial part only after an even number of edges. In particular, every cycle has even length.

F1
1.2

Conversely, suppose G has no odd cycle. Then by [L1] it has no odd closed walk.

givenL1
2.1

Let C be an arbitrary connected component. It is nonempty by [L2], so fix a root r∈C. Put AC equal to the vertices joined to r by an even-length path and BC equal to those joined to r by an odd-length path. Every vertex of C lies in at least one of these sets by [F2]. This construction is vacuous when the graph has no components.

step 1.2F2L2choose
3.1

The sets AC and BC are disjoint: an even r to v path followed by the reverse of an odd r to v path would be an odd closed walk, contrary to step 1.2.

step 1.2step 2.1
4.1

Every edge {u,v} inside C runs between AC and BC. Otherwise choose paths from r to u and from r to v having the common parity of their class. Traversing the first path, the edge {u,v}, and the reverse of the second path gives an odd closed walk, contrary to step 1.2.

step 1.2step 2.1step 3.1
5.1

By [L2], the connected components partition the finite vertex set and therefore form a finite family. Repeating this construction for each component and taking A and B to be the unions of their respective parts gives disjoint sets with V=A∪B. The endpoints of an edge are joined by that one-edge path and hence lie in one component by [F2], so every edge runs between the two global parts. Thus (A,B) is a bipartition.

step 2.1step 3.1step 4.1F1F2L2
6.1

Step 1.1 proves that bipartite graphs have no odd cycle, and steps 1.2 to 5.1 prove the converse.

step 1.1step 5.1∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices

Definition

On a finite vertex set V, the empty graph has edge set ∅ and the complete graph KV has edge set [V]2. When V is an n-element labelled set, these are also denoted K‾n and Kn.

For disjoint finite sets A and B, the complete bipartite graph KA,B has vertex set A∪B and edge set

{ {a,b}:a∈A, b∈B }.

If ∣A∣=m and ∣B∣=n, it is denoted Km,n.

The path graph Pn has n vertices. For n≥1, take vertices 0,…,n−1 and edges {i,i+1} for 0≤i<n−1; it has length n−1. Set P0 equal to the null graph. The cycle graph Cn is defined only for n≥3: it has vertices 0,…,n−1, the consecutive edges {i,i+1} for 0≤i<n−1, and the closing edge {n−1,0}. Thus in this notation the subscript records the number of vertices, not the length of a path.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

The Petersen graph on the two-element subsets of a five-element set, adjacent when disjoint

Definition

Fix a five-element set Ω. The Petersen graph has vertex set [Ω]2, the set of two-element subsets of Ω, and edge set

{ {A,B}⊆[Ω]2:∣{A,B}∣=2 and A∩B=∅ }.

In words, two vertices are adjacent exactly when the corresponding two-element subsets are disjoint. This is a finite simple graph, and its vertex set has (52)=10 elements by A finite set with n elements has exactly (n2) two-element subsets, and 2(n2)=n(n−1). The definition is independent of the names assigned to the five elements, because a relabelling preserves disjointness.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Graph isomorphisms, automorphisms and graph complements

Definition

Let G=(V,E) and H=(W,F) be graphs. A graph isomorphism is a bijection φ:V→W (Injection, surjection, bijection) such that, for all distinct u,v∈V,

{u,v}∈E⟺{φ(u),φ(v)}∈F.

Graphs are isomorphic, written G≅H, when such a map exists. An automorphism of G is an isomorphism from G to itself.

The complement of G is the graph

G‾:=(V,[V]2∖E).

Thus exactly one of G and G‾ contains any given pair of distinct vertices as an edge, and G‾‾=G.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Vertex and edge deletion, edge contraction, graph minors, subdivisions and topological minors

Definition

Let G=(V,E) be a graph. For v∈V, vertex deletion gives G−v:=G[V∖{v}]. For e∈E, edge deletion gives G−e:=(V,E∖{e}). The same notation is used for deleting a set of vertices or edges.

For an edge e={u,v}, contracting e identifies u and v. More precisely, take a new vertex xe, put V′=(V∖{u,v})∪{xe}, and let q:V→V′ send u,v to xe and fix every other vertex. Then

G/e:=(V′, { q[f]:f∈E∖{e}, ∣q[f]∣=2 }).

The condition ∣q[f]∣=2 deletes loops, while writing the edge family as a set merges parallel edges. Thus contraction returns a simple graph.

A graph H is a minor of G when it can be obtained by vertex deletions, edge deletions and edge contractions. Subdividing an edge {u,v} replaces it by {u,x} and {x,v} for a new vertex x. A subdivision repeats this operation zero or more times. The graph H is a topological minor of G when some subdivision of H is isomorphic to a subgraph of G (Subgraphs, induced subgraphs and spanning subgraphs).

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

The line graph whose vertices are the edges of the original graph and whose adjacency records a shared endpoint

Definition

For a graph G=(V,E), its line graph L(G) has vertex set E. Distinct vertices e,f∈E are adjacent in L(G) exactly when the edges e and f of G share an endpoint, that is, when e∩f≠∅.

Because E is finite and adjacency is recorded as a set of two-element subsets of E, the line graph is again a finite simple graph (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets). Incidence in G has become adjacency in L(G).

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Handshake lemma: the sum of the vertex degrees is twice the number of edges

Statement

For every finite simple graph G=(V,E),

∑v∈Vdeg⁡G(v)=2∣E∣

as an identity of natural numbers. It remains valid for the null graph, when both sides are 0.

Facts & Assumptions

Given: A finite simple graph G=(V,E).

[L1]

For a relation R⊆X×Y between finite sets, the sum of the row-fibre cardinalities and the sum of the column-fibre cardinalities both equal ∣R∣ (Double counting: ∑x∈X∣Rx∣=∣R∣=∑y∈Y∣Ry∣ for a relation between finite sets).

[F1]

The row and column fibres of a finite incidence relation are finite and are defined by fixing the first or second coordinate (A relation R⊆X×Y between finite sets, its row fibres Rx and its column fibres Ry).

[F2]

Proof

technique · direct
1.1

Define the incidence relation R:={ (v,e)∈V×E:v∈e }. Its row fibre at v is the set of edges incident with v, so it has cardinality deg⁡G(v) by [F2].

givenF1F2
1.2

Every edge e∈E is a two-element subset of V, so its column fibre has cardinality 2.

givenF1
2.1

Applying [L1] to R and using steps 1.1 and 1.2 gives ∑v∈Vdeg⁡G(v)=∣R∣=∑e∈E2=2∣E∣.

step 1.1step 1.2L1algebra∎
CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Every finite graph has an even number of vertices of odd degree

Statement

In every finite simple graph, the set of vertices having odd degree has even cardinality.

Facts & Assumptions

Given: A finite simple graph G=(V,E) and the finite set O:={v∈V:deg⁡G(v) is odd}.

[L1]

The handshake lemma gives ∑v∈Vdeg⁡G(v)=2∣E∣, an even natural number (Handshake lemma: the sum of the vertex degrees is twice the number of edges).

[F1]

The cardinality ∣O∣ is a natural number because O is a subset of the finite set V (The cardinality ∣A∣ of a finite set).

Proof

technique · direct
1.1

Split the degree sum into the vertices in O and those outside O. The latter summands are even, so their sum is even.

given
1.2

A finite sum of k odd natural numbers has the same parity as k: writing each summand as 2qi+1 makes the sum 2∑iqi+k. Hence ∑v∈Odeg⁡G(v) has the same parity as ∣O∣.

F1algebra
2.1

The total degree sum is even by [L1], and subtracting the even contribution from V∖O leaves an even sum over O. Step 1.2 therefore makes ∣O∣ even.

step 1.1step 1.2L1∎
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

The complete graph on an n-element vertex set has (n2) edges

Statement

If V is an n-element set, then the complete graph KV has exactly (n2) edges.

Facts & Assumptions

Given: A finite set V with ∣V∣=n.

[F1]

The edge set of KV is [V]2, the set of all two-element subsets of V (Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices).

Proof

technique · direct
1.1

By [F1], E(KV)=[V]2.

F1
2.1

By [L1], ∣[V]2∣=(n2), so step 1.1 gives ∣E(KV)∣=(n2).

step 1.1L1∎
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

There are 2(n2) simple graphs on a fixed labelled n-element vertex set

Statement

On a fixed labelled n-element vertex set V, there are exactly 2(n2) finite simple graphs.

Facts & Assumptions

Given: A fixed finite set V with ∣V∣=n.

[F1]

A simple graph on V is uniquely specified by choosing an edge set E⊆[V]2 (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets).

[L2]

A finite set with m elements has a power set with 2m elements (∣P(A)∣=2∣A∣ for finite A).

Proof

technique · direct
1.1

Sending a graph (V,E) to its edge set is a bijection from the simple graphs on the fixed labelled set V to the power set P([V]2), by [F1].

F1
2.1

By [L1] and [L2], ∣P([V]2)∣=2∣[V]2∣=2(n2). Combining with step 1.1 gives the stated count.

step 1.1L1L2∎
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Vertex cuts, edge cuts, vertex connectivity κ(G) and edge connectivity λ(G), with conventions for complete and one-vertex graphs

Definition

Let G=(V,E) be connected with n=∣V∣≥2. A vertex cut is a set S⊆V for which G−S is disconnected or has at most one vertex. An edge cut is a set F⊆E for which G−F is disconnected. The vertex connectivity and edge connectivity are

κ(G):=min⁡{ ∣S∣:S is a vertex cut },λ(G):=min⁡{ ∣F∣:F is an edge cut }.

The finite families being minimized are nonempty: deleting all but one vertex is a vertex cut, and deleting all edges is an edge cut. For the complete graph Kn, no deletion leaving at least two vertices disconnects it, so the first definition gives the standard exceptional convention κ(Kn)=n−1.

For the one-vertex graph K1, set κ(K1)=λ(K1)=0. The null graph and every disconnected graph are outside the domain of these two connectivity parameters under this convention. Thus the inequalities involving δ(G) are stated only for nontrivial connected graphs, where the vertex set is nonempty and δ(G) is defined.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

A bridge as an edge whose deletion increases the number of connected components

Definition

Let G be a finite graph and let e be an edge. Using edge deletion from Vertex and edge deletion, edge contraction, graph minors, subdivisions and topological minors and connected components from Connected graphs and connected components defined by the existence of vertex paths, the edge e is a bridge if G−e has more connected components than G. Component vertex sets partition the graph as in The connected components of a graph partition its vertex set and are its maximal connected subgraphs, so this comparison is unambiguous even when G is disconnected.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

An edge of a finite graph is a bridge if and only if it lies on no cycle

Facts & Assumptions

Given: A finite graph G and an edge e=uv.

[F1]

A bridge is an edge whose deletion increases the number of connected components (A bridge as an edge whose deletion increases the number of connected components).

[F2]

A cycle is a closed walk with no repeated vertices apart from its first and last vertex (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

Proof

technique · direct
1.1

If e lies on a cycle, the remaining edges of that cycle form a u-v path in G−e. Every path in G that used e can replace that occurrence by this path, so deleting e separates no formerly connected pair. Thus e is not a bridge.

F1F2
1.2

Conversely, if e is not a bridge, u and v remain in the same component of G−e and hence are joined there by a path. Adding e to that path gives a cycle containing e.

F1F2
2.1

Step 1.1 says an edge on a cycle is not a bridge, and step 1.2 says an edge not a bridge lies on a cycle. Taking the contrapositive of either implication and combining them proves the biconditional componentwise, including when G is disconnected.

step 1.1step 1.2∎
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

For every nontrivial connected graph, λ(G)≤δ(G)

Statement

For every connected finite simple graph G with at least two vertices, λ(G)≤δ(G).

Facts & Assumptions

Given: A connected graph G=(V,E) with ∣V∣≥2.

[F2]

A vertex v of minimum degree has exactly δ(G) incident edges (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

Proof

technique · direct
1.1

Choose a vertex v with deg⁡G(v)=δ(G) and let F be the set of all edges incident with v. Then ∣F∣=δ(G).

givenF2choose
2.1

In G−F, the vertex v is isolated while at least one other vertex remains, so G−F is disconnected. Thus F is an edge cut.

step 1.1F1
3.1

Minimality in [F1] gives λ(G)≤∣F∣=δ(G).

step 1.1step 2.1F1∎
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)Open item page →

For every nontrivial connected graph, κ(G)≤λ(G)

Statement

For every connected finite simple graph G with at least two vertices, κ(G)≤λ(G).

Facts & Assumptions

Given: A connected graph G=(V,E) with n=∣V∣≥2.

[F1]

κ(G) is the least size of a vertex set whose deletion disconnects G or leaves at most one vertex, and λ(G) is the least size of an edge cut (Vertex cuts, edge cuts, vertex connectivity κ(G) and edge connectivity λ(G), with conventions for complete and one-vertex graphs).

Proof

technique · constructive
1.1

Choose an edge cut F of size λ(G). Let A be the vertex set of one component of G−F and put B=V∖A. Then A and B are nonempty, and every edge of G between A and B belongs to F.

givenF1F2chooseconstruct
2.1

If ∣F∣≥n−1, deleting all but one vertex gives a vertex cut of size n−1≤∣F∣, so κ(G)≤λ(G).

step 1.1F1
2.2

Suppose ∣F∣<n−1. The bipartite set of all pairs with one endpoint in A and one in B has ∣A∣∣B∣≥∣A∣+∣B∣−1=n−1 members. Hence not every such pair is an edge of G, for all cross edges lie in F. Choose nonadjacent a∈A and b∈B.

step 1.1algebrachoose
3.1

For each edge e∈F, choose one endpoint s(e) different from both a and b; this is possible because e≠{a,b}. Put S={s(e):e∈F}. Then a,b∉S, every edge of F meets S, and ∣S∣≤∣F∣.

step 2.2construct
4.1

Any path from a∈A to b∈B must use an edge between A and B, hence an edge of F by step 1.1, and therefore a vertex of S by step 3.1. Thus G−S has no a to b path and is disconnected, so S is a vertex cut.

step 1.1step 3.1F1F2
5.1

In the case ∣F∣<n−1, step 4.1 gives κ(G)≤∣S∣≤∣F∣=λ(G); step 2.1 gives the same inequality in the remaining case.

step 2.1step 3.1step 4.1F1discharge-construct∎
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Whitney's inequalities: κ(G)≤λ(G)≤δ(G) for every nontrivial connected graph

Statement

For every connected finite simple graph G with at least two vertices,

κ(G)≤λ(G)≤δ(G).

This includes complete graphs under the convention κ(Kn)=n−1: for Kn with n≥2, all three quantities equal n−1.

Facts & Assumptions

Given: A connected finite simple graph G with at least two vertices.

[L1]

κ(G)≤λ(G) for every nontrivial connected graph (For every nontrivial connected graph, κ(G)≤λ(G)).

[L2]

λ(G)≤δ(G) for every nontrivial connected graph (For every nontrivial connected graph, λ(G)≤δ(G)).

Proof

technique · direct
1.1

Applying [L1] and [L2] to G gives κ(G)≤λ(G)≤δ(G).

L1L2
2.1

For Kn, deleting fewer than n−1 vertices leaves a nonempty complete graph and deleting n−1 leaves one vertex, so κ(Kn)=n−1; every vertex has degree n−1, and deleting all n−1 edges incident with one vertex is an edge cut, while any smaller edge deletion leaves every pair joined through a remaining direct edge or a two-edge path. Hence λ(Kn)=δ(Kn)=n−1, as stated.

L1L2algebra∎
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Every nonempty finite simple graph contains a path with at least δ(G) edges

Statement

Every nonempty finite simple graph G contains a path having at least δ(G) edges.

Facts & Assumptions

Given: A nonempty finite simple graph G=(V,E).

[F2]

A path has no repeated vertex, and appending a neighbour not already on it produces a longer path (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

[F3]

A finite graph has only finitely many paths, so a path of greatest length exists (The cardinality ∣A∣ of a finite set).

Proof

technique · constructive
1.1

Choose a path P=(v0,v1,…,vℓ) of greatest length in G.

givenF3chooseconstruct
2.1

Every neighbour of v0 lies among v1,…,vℓ: a neighbour outside this list could be prepended to P, giving a longer path and contradicting step 1.1.

step 1.1F2
3.1

The deg⁡G(v0) distinct neighbours of v0 therefore occupy distinct positions among the ℓ vertices v1,…,vℓ, so ℓ≥deg⁡G(v0)≥δ(G) by [F1]. Thus P has at least δ(G) edges.

step 2.1F1discharge-construct∎
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Every finite simple graph of minimum degree at least two contains a cycle

Statement

If G is a finite simple graph with δ(G)≥2, then G contains a cycle.

Facts & Assumptions

Given: A finite simple graph G with δ(G)≥2.

[L1]
[F1]

A closed walk of length at least 3 whose vertices before the final repeated endpoint are distinct is a cycle (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

Proof

technique · direct
1.1

By [L1], G has a path of length at least 2. Among all paths choose one of greatest length, say P=(v0,…,vℓ), so ℓ≥2.

givenL1choose
2.1

Every neighbour of v0 lies on P, since a neighbour outside P could be prepended to make a longer path. Since deg⁡G(v0)≥2, besides v1 there is a neighbour vj with j≥2.

step 1.1given
3.1

The list (v0,v1,…,vj,v0) is a closed walk of length j+1≥3 with distinct vertices before its final v0, so it is a cycle by [F1].

step 1.1step 2.1F1∎
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)Open item page →

An n-vertex simple graph with more than (n−12) edges is connected

Statement

Let G be a finite simple graph with n≥1 vertices. If ∣E(G)∣>(n−12), then G is connected.

Facts & Assumptions

Proof

technique · contradiction
1.1

Suppose, for contradiction, that G is disconnected. Choose one component with vertex set A, put B=V∖A, and write k=∣A∣. Then 1≤k≤n−1, ∣B∣=n−k, and no edge joins A to B.

assume-contraF1choose
2.1

By [L1], the edges with both endpoints in A number at most (k2), and those with both endpoints in B number at most (n−k2). Step 1.1 accounts for every edge, so ∣E∣≤(k2)+(n−k2).

step 1.1L1
2.2

For 1≤k≤n−1, elementary binomial algebra gives (k2)+(n−k2)=(n−12)−(k−1)(n−k−1)≤(n−12).

step 1.1algebra
3.1

Steps 2.1 and 2.2 give ∣E∣≤(n−12), contradicting the hypothesis. Therefore G is connected.

step 2.1step 2.2givendischarge-contradiction∎

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

FALSE: a finite simple graph is determined up to isomorphism by its degree sequence

Statement

FALSE. A finite simple graph is determined up to isomorphism by its degree sequence.

Facts & Assumptions

Given: The cycle graph C6 and the graph H that is the disjoint union of two copies of C3.

[F2]

A graph isomorphism is a bijection preserving adjacency in both directions, and therefore sends cycles to cycles of the same length (Graph isomorphisms, automorphisms and graph complements).

Refutation

technique · direct
1.1

Every vertex of C6 has degree 2. Every vertex of H lies on one of its two triangles and also has degree 2. Thus both degree sequences are (2,2,2,2,2,2).

givenF1F3
1.2

The graph H contains a cycle of length 3, namely either triangle, whereas C6 contains no triangle because its only edges join consecutive vertices on its six-cycle.

givenF1
2.1

If C6≅H, [F2] would carry a triangle of H to a triangle of C6, contradicting step 1.2. Hence the graphs have the same degree sequence but are not isomorphic, refuting the claim.

step 1.1step 1.2F2∎
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)Open item page →

FALSE: a connected simple graph in which every vertex has even degree must itself be a cycle

Statement

FALSE. Every connected finite simple graph in which every vertex has even degree is a cycle graph.

Facts & Assumptions

Given: The bowtie graph with vertices x,a,b,c,d and edges xa,ab,bx,xc,cd,dx, consisting of two triangles that share only x.

[F2]

A connected graph has a path between every two vertices (Connected graphs and connected components defined by the existence of vertex paths).

[F3]

In the cycle graph Cn each vertex is joined by an edge exactly to its predecessor and its successor modulo n (Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices); with [F1] this gives every vertex of a cycle graph exactly two neighbours.

Refutation

technique · direct
1.1

Every two vertices of the bowtie are joined by a path, using x when they lie in different triangles, so the graph is connected by [F2].

givenF2
1.2

The vertex x has neighbours a,b,c,d and hence degree 4; each other vertex has degree 2. Thus all degrees are even by [F1].

givenF1
2.1

The bowtie is not a cycle graph because its vertex x has degree 4, whereas [F3] gives degree 2 at every vertex of a cycle graph. Steps 1.1 and 1.2 therefore refute the claim.

step 1.1step 1.2F3∎
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

FALSE: vertex connectivity, edge connectivity and minimum degree are always equal

Statement

FALSE. Every nontrivial connected finite simple graph satisfies κ(G)=λ(G)=δ(G).

Facts & Assumptions

Given: The bowtie graph G with two triangles xabx and xcdx sharing only the vertex x.

[L1]

Whitney's theorem guarantees only κ(G)≤λ(G)≤δ(G) (Whitney's inequalities: κ(G)≤λ(G)≤δ(G) for every nontrivial connected graph).

Refutation

technique · direct
1.1

Deleting x separates the two edges ab and cd, while the connected graph has no vertex cut of size 0. Hence κ(G)=1.

givenF1
1.2

Deleting the two edges xa and xb separates a,b from the other triangle, so λ(G)≤2. No single edge disconnects G, because every edge lies on one of the two triangles and the other two edges of that triangle give an alternate path between its endpoints. Hence λ(G)=2.

givenF1
1.3

The four noncentral vertices have degree 2 and x has degree 4, so δ(G)=2.

given
2.1

Thus κ(G)=1<2=λ(G)=δ(G), a strict instance of the first Whitney inequality and a counterexample to the asserted equality.

step 1.1step 1.2step 1.3L1∎
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

FALSE: a shortest walk between two vertices may repeat a vertex and fail to be a path

Statement

FALSE. A shortest walk between two vertices may repeat a vertex and fail to be a path.

Facts & Assumptions

Given: Vertices u,v in one connected component and a u to v walk W of minimum length.

[L1]

Every walk contains a path between the same endpoints of no greater length, and if the walk repeats a vertex the path can be chosen strictly shorter (Every walk between two vertices contains a path between the same endpoints).

[F1]

The distance dG(u,v) is the minimum length of a u to v path (Graph distance within a component, eccentricity, diameter and girth, including the acyclic convention).

Refutation

technique · contradiction
1.1

Suppose, for contradiction, that the minimum-length walk W repeats a vertex and is not a path.

assume-contragiven
2.1

By [L1], deleting a closed segment from W yields a u to v path of length strictly less than the length of W. This path is also a walk, contradicting the minimality of W.

step 1.1L1F1
3.1

Therefore every shortest walk is a path, and the claimed possibility cannot occur.

step 2.1discharge-contradiction∎

Sources