Lemma (Bellman Criteria) [Brandes 2001]
Let $G$ be a graph and $s, t, v \in V(G)$.
A vertex $v \in V(G)$ lies on a shortest path between vertices $s, t \in V$, if and only if dist(s,t) = dist(s,u)+dist(u,t).
Can i show by the definition that dist(s,v)+dist(v,t) is the shortest path from $s$ to $t$ via $v$ and then show by contradiction that ">" "<" is valid and thus equality must apply? Or is it possible to show that only by equivalence direction?
The distance between two vertices is by definition the length of a shortest path between them. So if $d(s,v)+d(v,t)=d(s,t)$ there is an $st$-path containing $v$, which is a shortest path from $s$ to $t$.
Conversely assume that $d(s,v)+d(v,t)\neq d(s,t)$ and note that by definition of the latter distance we indeed have $d(s,v)+d(v,t)>d(s,t)$. Moreover any $st$-path $P$ containing $v$ satisfies $\operatorname{len} P \geq d(s,v) + d(v,t)$, since it decomposes into an $sv$-path $P_s$ and $vt$-path $P_t$, which by definition satisfy $\operatorname{len} P_s \geq d(s,v)$ and $\operatorname{len} P_t \geq d(v,t)$. But this way $v$ cannot possibly lie on a shortest $st$-path.