Filtrator is a pair $(\mathfrak A; \mathfrak Z)$ of a poset $\mathfrak A$ and it subset $\mathfrak Z$.
Let $a\in\mathfrak A$. Then by definition $\operatorname{up} a = \{x\in\mathfrak Z \mid x\ge a \}$.
Prefiltered filtrator is such a filtrator that $\operatorname{up}$ is injective.
Semifiltered filtrator is such a filtrator that $\forall a,b\in\mathfrak A: (\operatorname{up} a\supseteq\operatorname{up} b \Rightarrow a\le b)$.
It is obvious that every semifiltered filtrator is prefiltered.
Is every prefiltered filtrator semifiltered?
No. Take $\mathfrak{A} = \{a,b\}$ with $\leq$ being equality and $\mathfrak{Z} = \{b\}$. Then $\operatorname{up} a = \emptyset \subseteq \{b\} = \operatorname{up} b$, so $\operatorname{up}$ is injective, hence the filtrator is prefiltered, but because of $a \not\leq b$ the filtrator is not semifiltered.