Hypergraph Variant of Handshaking Lemma The handshaking lemma for $2$-graphs says that for any simple graph $G$
$$
2e(G) = \sum_{v \in G} d(v).
$$
Is there a generalization of this fact to $r$-uniform hypergraphs for $r>2$ ?
Let's compute the degree sum and see what happens:
$$
\sum_{v \in G} d(v) = \sum_{v \in G} |\{e \in E(G) \ : \ v \in e \}|
$$
How many times are we counting each edge? Well, each edge is being counted $r$ times,
once for each $v \in e$ (not sure about this reasoning here). Thus we must have that
$$
\frac{1}{r}\sum_{v \in G} d(v) = e(G)
$$
Here is a possible justification for this using an auxillary bipartite graph.
Let $H$ be a bipartite graph with partitions $A,B$ such that $A = V(G)$ and
$B = E(G)$ with adjacency defined as follows: for each $v \in A$ and $e \in B$ we have
$$
v \sim e \iff v \in e.
$$
Counting the edges of $H$ from the perspective of each partition gives
$$
 r\cdot e(G)= \sum_{b \in B}d_H(b) = e(H) = \sum_{a\in A} d_H(a) = \sum_{v \in V(G)} d_G(v).
$$
(Slightly meta follow up question: If my reasoning is correct then why is this more generalized form of the handshaking lemma not publicized more? Maybe my search queries are not articulated well enough.)
 A: It's true that for a hypergraph $H=(V,E)$ with the flag multiset $F$, $\sum\limits_{v\in V}deg(v) = \sum\limits_{e\in E}|e| = |F|$. As it was pointed out in the comments, double counting is one of the names you may encounter (especially so in extremal combinatorics) referring to the technique used to proof this identity (e.g. here).
The reason why you might have troubles finding this result is that, apart from some classics (e.g. Claude Berge's "Hypergraphs, Combinatorics of finite sets"), you won't find many works treating the hypergprah theory in general that would point out this (and other, less trivial) results. However if you read papers on extremal graph theory or something that deals with counting problems, you will sometimes see authors proving this lemma just to get over it on their way to a more glorious goal.
A: Copy and pasted from original post:
Let's compute the degree sum and see what happens:
$$
\sum_{v \in G} d(v) = \sum_{v \in G} |\{e \in E(G) \ : \ v \in e \}|
$$
How many times are we counting each edge? Well, each edge is being counted $r$ times,
once for each $v \in e$ (not sure about this reasoning here). Thus we must have that
$$
\frac{1}{r}\sum_{v \in G} d(v) = e(G)
$$
Here is a possible justification for this using an auxillary bipartite graph.
Let $H$ be a bipartite graph with partitions $A,B$ such that $A = V(G)$ and
$B = E(G)$ with adjacency defined as follows: for each $v \in A$ and $e \in B$ we have
$$
v \sim e \iff v \in e.
$$
Counting the edges of $H$ from the perspective of each partition gives
$$
 r\cdot e(G)= \sum_{b \in B}d_H(b) = e(H) = \sum_{a\in A} d_H(a) = \sum_{v \in V(G)} d_G(v).
$$
