2. Proof:
Let's say ? is the constant of the pumping lemma, and let's consider ? = ??N?N ∈ ? and |?| = 2?+1 ≥ ?. By the lemma, it can be written ? = ???, such that |??| ≤ ? with ? ≠ ? such that for all ?≥0 it is ??k? ∈ ?.
Now ? must be composed of only one symbol (in this case, as |??|≤?, it has to be only ? or some ?s) as otherwise ??k? for ?>1 isn't of the right form (?s followed by ?s followed by ?s). Now ?=?, ?=? is impossible, because in that case we have ??0? = ?N?N ∉ ?. But ? just ?s with ?≠1 gives a string with number of ?s and ?s different, thus not in ?. No division into ?,?,? works, the language fails the pumping lemma. Hence it can't be regular.
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