Is $\Pi’\propto_{poly}\Pi$ for any two NP-complete problems $\Pi,\Pi'$? I am facing the following question:

Let $\Pi$ and $\Pi$’ be two NP-complete problems, prove or refute $\Pi’\propto_{poly}\Pi$.

I do not understand the meaning of this question and how to answer it.
The definition of $\Pi’\propto_{poly}\Pi$&quot; as follows:

Let $\Pi$ and $\Pi'$ be two decision problems. We say that $\Pi$ reduces to $\Pi'$ in polynomial time, symbolized as $\Pi’\propto_{poly}\Pi$, if there exists a deterministic algorithm $A$ that behaves as follows. When $A$ is presented with an instance $I$ of problem $\Pi$, it transforms it into an instance $I’$ of problem $\Pi'$ such that the answer to $I$ is yes iff the answer to $I’$ is yes. Moreover, this transformation must be achieved in polynomial time.
Try to combine the following two definitions:

A problem $\Pi$ is NP-hard if $\Pi' \propto_{\mathit{poly}} \Pi$ for every $\Pi'$ in NP.
A problem $\Pi$ is NP-complete if $\Pi$ is NP-hard and $\Pi$ is in NP.