Final answer:
Given regular expressions r and s, where L(r) is a subset of L(s), the expression a = ⟨r s⟩ is decidable. The language generated by a is regular, proving that a is decidable.
Step-by-step explanation:
The statement claims that given regular expressions r and s, where the language generated by r is a subset of the language generated by s, the expression a = ⟨r s⟩ is decidable. To prove this, we need to show that one of the given options (A, B, C, or D) is true. Let's evaluate each option:
A) The complement of a is recursively enumerable: This option is unrelated to the decidability of a, so it does not help in proving that a is decidable.
B) The intersection of L(a) (the language generated by a) and L(s) is empty: Since we know that L(r) ⊆ L(s), and a = ⟨r s⟩, it follows that L(a) ⊆ L(s). Therefore, the intersection of L(a) and L(s) will always be non-empty, making this option false.
C) The language generated by a is regular: We know that the concatenation of regular languages is also regular. Since r and s are regular expressions, their concatenation a = ⟨r s⟩ is also regular. Therefore, this option is true and proves that a is decidable.
D) The concatenation of r and s is context-free: This option does not help in proving the decidability of a. It is unrelated to the regularity of the language generated by a. Therefore, it is false and does not contribute to proving that a is decidable.
Based on the analysis above, option C is the correct answer, proving that a is decidable.