Final answer:
0° 1ᵇ # 0ᶜ 1ᵈ € B if and only if the string contains an equal number of 0's and 1's before and after the # symbol. To prove that B is not context-free, we can use the Pumping Lemma and consider two cases based on whether the pumping substring includes the # symbol or not. F is context-free as it can be represented by a context-free grammar.
Step-by-step explanation:
(a) To complete the statement, we need to find the condition that makes 0° 1ᵇ # 0ᶜ 1ᵈ € B true. This means that the string formed by concatenating the 0's and 1's must appear before and after the # symbol, and the exponents b and d can be any positive integers. So, the statement is true if the string contains an equal number of 0's and 1's before and after the # symbol.
(b) To prove that B is not context-free using the Pumping Lemma, we can break into two cases: when the pumping substring vxy includes the # symbol, and when the pumping substring vxy does not include the # symbol. In both cases, we can find a string in B that cannot be pumped, proving that B is not context-free.
(c) To prove that F is context-free or not, we need to analyze its structure. F is a language that consists of a string w followed by a # symbol and another string q, where both w and q are formed by 0's and 1's. Since F can be represented by a context-free grammar, it is context-free.