Question

Let A = {a'bic*:0≤i≤ j ≤ k}. Use the pumping lemma for context-free languages to prove that A is not a context-free language.

Answer 1

To prove that A is not a context-free language, we can use the pumping lemma for context-free languages. Let's assume that A is a context-free language. According to the pumping lemma, there exists a pumping length p, such that for any string s in A with a length of at least p, s can be divided into five parts: uvwxy, satisfying three conditions.

Step-by-step explanation:

To prove that A is not a context-free language, we can use the pumping lemma for context-free languages.

Let's assume that A is a context-free language. According to the pumping lemma, there exists a pumping length p, such that for any string s in A with a length of at least p, s can be divided into five parts: uvwxy, satisfying three conditions:

1. vwx has a length less than or equal to p
2. For any non-negative integer i, the string uv^iwx^iy is also in A
3. If uvwxy contains any non-terminal symbols, then we can assume that vwx contains at most one non-terminal symbol

Now, let's choose the string s = a^pb^pc^pd^p. This string belongs to A because we have 0 ≤ i ≤ j ≤ k, and there are non-terminal symbols in the string. However, if we choose i = 0, the resulting string will have a different number of a's, b's, c's, and d's, which means it will no longer belong to A. This contradicts the pumping lemma, proving that A is not a context-free language.