Prove that the following languages are not context-free.

i) L = {a^i | i is prime}
ii) L = {a^n | n ≥ 1}

The language L = {a^i | i is prime} is not context-free because it can be shown using the Pumping Lemma that pumping parts of strings can lead to non-prime counts. The second language mentioned, L = {a^n | n ≥ 1}, is actually context-free as it generates all strings of one or more 'a's. Therefore, only the first language is proven not to be context-free.
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Answered by Anonymous | 2025-07-04

To demonstrate that a language is not context-free, one way is to use the Pumping Lemma for context-free languages. The Pumping Lemma provides a condition that all context-free languages must satisfy. If we can show that a particular language does not satisfy this condition, then the language is not context-free.
Let's address each language separately:
i) Language L = {a^i | i \text{ is prime}}

Pumping Lemma for Context-Free Languages: According to this lemma, if a language L is context-free, there exists some integer p (pumping length) such that any string s in this language with length at least p can be divided into five parts, s = uvwxy, satisfying the following conditions:

The string vwx has a length of at most p.
The string vx is not empty (i.e., |vx| > 0).
For all integers n ≥ 0, the string u(v^n)w(x^n)y is in L.


Application to L = {a^i | i \text{ is prime}}: Let's assume L is context-free. Then, by the Pumping Lemma, there exists a pumping length p.
Consider the string s = a^q, where q is a prime number greater than p. According to the Pumping Lemma, s can be split into uvwxy, with the conditions mentioned above.
If we "pump" the string, creating strings u(v^n)w(x^n)y, for n > 2, we see that the resulting strings will be in the form a^(q+m) for some integer m > 0.
If n is increased, the new string cannot be guaranteed to have a prime number of a's, violating the condition that the string must belong to L. Therefore, no such pumping length p can exist, showing a contradiction, and that means L is not context-free.


ii) Language L = {a^n | n \geq 1}

Observation of Language Nature: This language is simply strings consisting of any number of a’s (such as "a", "aa", "aaa", etc.), which actually forms a context-free language.

Contradiction in the Problem Statement: The problem seems to suggest proving that this language is not context-free. However, {a^n | n \geq 1} is context-free since it can be generated by the context-free grammar:

S -> aS | a



Hence, there is no need for a proof that this language is not context-free because it actually is context-free. The task of proving it is not context-free is a misunderstanding.
By these analyses, we confirm that the first language is not context-free, while the second one is context-free.

Answered by MasonWilliamTurner | 2025-07-06