The core of this problem is understanding that a language is Recursively Enumerable (RE) if and only if a recognizer (a Turing Machine) can be built for it. A recognizer must halt and accept any string that is in the language. Let's analyze whether we can build such a recognizer for $L_1$ and $L_2$
Step 1: Analyze the base sets
- $S_{\text {hall }}$ : This is the language for the Halting Problem on a fixed input $\epsilon$. This is the canonical example of a language that is Recursively Enumerable (RE) but not Recursive. It is RE because we can recognize it by simulating the machine $M$ on $\epsilon$; if it halts, we accept.
- $S_{\text {loop }}$ : This is the language for the Looping Problem. It is the complement of $S_{\text {halt }}$. The complement of an RE-but-not-Recursive language is not Recursively Enumerable. There is no general algorithm that can confirm in finite time that a machine will loop forever.
Step 2: Analyze Language $L_1$
$L_1=\left\{\left\langle M^{\prime}\right\rangle \mid L\left(M^{\prime}\right) \cap S_{\text {halt }} \neq \emptyset\right\}$. Can we build a recognizer for $L_1$ ?
Yes. A recognizer for $L_1$ takes an input $\left\langle M^{\prime}\right\rangle$ and must determine if $M^{\prime}$ accepts any string $w$ which happens to be in $S_{\text {halt }}$. The recognizer can perform the following procedure using dovetailing (parallel simulation):
1. Enumerate all possible strings $w_1, w_2, w_3, \ldots$ over the input alphabet.
2. For each string $w_i$ :
- Task A: Check if $w_i \in L\left(M^{\prime}\right)$. To do this, simulate $M^{\prime}$ on input $w_i$.
- Task B: Check if $w_i \in S_{\text {halt }}$. To do this, verify $w_i$ is a valid encoding $\left\langle M_{w_i}\right\rangle$ and then simulate $M_{w_i}$ on input $\epsilon$.
3. The recognizer runs all instances of Task $A$ and Task $B$ in parallel.
4. If for any string $w_i$, both its Task A and its Task B halt and accept, then the recognizer has found a string that is both accepted by $M^{\prime}$ and is in $S_{\text {hall }}$. The recognizer immediately halts and accepts $\left\langle M^{\prime}\right\rangle$.
Since $S_{\text {halt }}$ is RE, Task B is a valid semi-decision procedure. Task A is also a valid semi-decision procedure. If $\left\langle M^{\prime}\right\rangle \in L_1$, such a string $w_i$ exists, and this process is guaranteed to find it and accept. Therefore, $L_1$ is Recursively Enumerable.
Step 3: Analyze Language $L_2$
$L_2=\left\{\left\langle M^{\prime}\right\rangle \mid L\left(M^{\prime}\right) \cap S_{\text {loop }} \neq \emptyset\right\}$. Can we build a recognizer for $L_2$ ?
No. Let's attempt to use the same strategy. The recognizer would need to find a string $w$ such that $M^{\prime}$ accepts $w$ AND $w \in S_{\text {loop }}$.
- The task of verifying " $M^{\prime}$ accepts $w$ " is semi-decidable.
- The task of verifying " $w \in S_{\text {loop }}$ " (i.e., the TM encoded by $w$ loops on $\epsilon$ ) is not semidecidable, because $S_{\text {loop }}$ is not an RE language.
There is no sub-procedure that can run and eventually halt to confirm " $w \in S_{\text {loop }}$ ". A recognizer for $L_2$ would need such a confirmation. Since one of its fundamental sub-tasks is not semidecidable, we cannot construct a recognizer for $L_2$. Therefore, $L_2$ is not Recursively Enumerable.
This makes option (B) the correct statement.
