Let $X$ be a recursive language and $Y$ be a recursively enumerable but not recursive language. Let $W$ and $Z$ be two languages such that $\overline{Y}$ reduces to $W$, and $Z$ reduces to $\overline{X}$ (reduction means the standard many-one reduction). Which one of the following statements is TRUE?

Given:

• $X$ is recursive (decidable). Since recursive languages are closed under complementation, $\overline{X}$ is also recursive.

• $Y$ is recursively enumerable (RE) but not recursive (not co-RE). Therefore, its complement $\overline{Y}$ is not recursively enumerable (not RE).

Now consider the reductions ($A \le_m B$):

1. For $W$ ($\overline{Y} \le_m W$):

   • If $W$ were recursively enumerable, then by reduction properties, $\overline{Y}$ would also have to be recursively enumerable.

   • However, we know $\overline{Y}$ is not RE.

   • Hence, $W$ cannot be recursively enumerable ($W$ is not RE).

2. For $Z$ ($Z \le_m \overline{X}$):

   • Since $\overline{X}$ is recursive (decidable), and any language that reduces to a recursive language is itself recursive, $Z$ must be recursive.

Therefore, $W$ is not recursively enumerable and $Z$ is recursive.

This matches option (C).