Final answer:
To show that the class of context-free languages is not closed under the star operation, we can use a counter-example with the language L=w^2. Taking w to be a string generated by a context-free grammar G, we can show that L is not context-free using the pumping lemma for context-free languages. Therefore, option a) fails to prove the closure of context-free languages under star.
Step-by-step explanation:
To show that the class of context-free languages is not closed under the star operation, we need to find a counter-example. Let's consider the language A=w^k, where w is a string generated by a context-free grammar G and k≥0.
If we take the language L=w^2, where w∈A, we can see that L is not context-free. To prove this, we can use the pumping lemma for context-free languages which states that if L is context-free, there exists a constant p such that any string s in L with |s|≥p can be split into five parts u, v, x, y, and z, satisfying the conditions: s=uvxyz, |vxy|≤p, |vy|≥1, and for all i≥0, the string uvikxykz is also in L. If we choose the string s=w2∈L, we will find that it cannot be split according to the conditions of the lemma, thus proving that L is not context-free.
Therefore, option a) fails to prove that the class of context-free languages is closed under star since the language L=w^2, where w∈A, is not context-free.