Getting inequality from statement I have read in book of combinatorics this statement: > any vertex is incidental with at least four edges, and each edge is incidental with two vertices, so $$ 4 |V| \le 2 |E| $$ where V is set of vertices and E set of edges I suspect that it is easy but... how I can get this inequality from this statement in formal way, not just write and say "it is obvious because it follows from statement". > any vertex is incidental with at least four edges So $$\forall_v \exists_{e_1, e_2, e_3, e_4, (e_i \neq e_j)} \forall_{k \in \left\\{1,2,3,4 \right\\}} v \in e_k$$ $$|V| \le 4|E|$$ > each edge is incidental with two vertices $$ |V| = \frac{1}{2}|E| $$ $$ 2|V| = |E| $$ but I don't get result :( Can somebody explain me, how they get this inequality?

Let us count the elements of the set $$X=\\{\,(v,e)\mid v\in e\,\\}$$ in two different ways.

The canonical projection map $\pi_E\colon X\to E$ is 2-to-1. That means that each element of $E$ has _exactly_ two pre-images, i.e., $|X|=2|E|$. On the other hand, with the canonical projection map $\pi_V\colon X\to V$, each element of $V$ has _at least_ four pre-images. If you don't see immediately that his implies $|X|\ge 4|V|$, pick $4$ pre-images for each $v\in V$ and thereby find a subset $X'\subset X$ such that $\pi_V$ restricted to $X'$ is 4-to-1, hence exactly $|X'|=4|V|$. Putting things together, $$ 4|V|=|X'|\le |X|=2|E|.$$