TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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

Statement

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

$\sum_{v\in V}\deg_G(v)=2|E|$

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 $R\subseteq X\times Y$ between finite sets, the sum of the row-fibre cardinalities and the sum of the column-fibre cardinalities both equal $|R|$ (Double counting: $\sum_{x \in X}\lvert R_x\rvert = \lvert R\rvert = \sum_{y \in Y}\lvert R^y\rvert$ 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 $R \subseteq X \times Y$ between finite sets, its row fibres $R_x$ and its column fibres $R^y$ ).

[F2]

$\deg_G(v)$ is the number of edges incident with $v$ (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

Proof

technique · direct

1.1

Define the incidence relation $R:=\{\,(v,e)\in V\times E:v\in e\,\}$ . 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 $e\in E$ 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 $\sum_{v\in V}\deg_G(v)=|R|=\sum_{e\in E}2=2|E|$ .

step 1.1step 1.2L1algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Cornell Mathematics, Graph Theory notes (standard reference, not scraped)