There is a conjecture on $P\overset{?}{=}PSPACE$ in computational complexity zoo. It is a widespread belief that $P\neq PSPACE$, otherwise $P=NP$ which is extremely impossible. In this short work, we assert that $P\neq PSPACE$ no matter what outcome is on $P\overset{?}{=}NP$. We accomplishe this via showing $NP\neq PSPACE$. The method is by the result that Circuit-SAT is $\leq_{\log}$--complete for $NP$, Circuit-SAT$\in DSPACE[n]$, the known result $DSPACE[n^{1+c}]\subset DSPACE[n^{2(1+c)}]$ (indicated by the space complexity hierarchy theorem) and the fact that $PSPACE$ is the union set of all $DSPACE[n^k]$ where $k\in\mathbb{N}$. Closely related consequences are summarized.

翻译：在计算复杂动物园中有一个关于$P\ overset{?\\\\\\\\\\\\\\\\\\\\\PSPACE$的猜测。 一种普遍的看法是, $P\neq PSPACE$, 或者说, $P=NP$, 这是极不可能的。 在这一短短的工程中, 我们声称, $P\neq PSPACE$, 不管$P\ overset{\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\