Given two languages : \(L_1 = \{(ab)^n a^k \mid n \gt k,\ k \ge 0\}\) \(L_2 = \{a^n b^m \mid n \ne m\}\) Using pumping lemma for regular language, it can be shown that
L1 is not regular and L2 is not regular.
Pumping lemma: if L is regular, there is a p such that every w ∈ L with |w| ≥ p splits as w = xyz with |xy| ≤ p, |y| ≥ 1 and \(xy^i z \in L\) for all i ≥ 0. Finding one w for which every split fails proves L is not regular.
For \(L_1\) take \(w = (ab)^p a^{p-1}\) (n = p > k = p – 1). Since |xy| ≤ p, y lies inside the (ab) part. Pump down (i = 0): if |y| is even, deleting it removes |y|/2 whole ab-blocks, leaving \((ab)^{p-j} a^{p-1}\) with p – j ≤ p – 1, which breaks n > k; if |y| is odd, deleting it puts two equal letters together or makes the string start with b, which no string of \(L_1\) allows. So \(L_1\) is not regular.
For \(L_2\) take \(w = a^p b^{p+p!}\). Here y = \(a^j\) with 1 ≤ j ≤ p, so j divides p!. Pump with \(i = 1 + \frac{p!}{j}\): the number of a's becomes \(p + p!\), equal to the number of b's, so the string leaves \(L_2\). Hence \(L_2\) is not regular (equivalently, \(\overline{L_2} \cap a^*b^* = \{a^n b^n\}\) is not regular).
Both languages need unbounded counting, which a finite automaton cannot do: \(L_1\) is not regular and \(L_2\) is not regular.
Hence, the answer is L1 is not regular and L2 is not regular..
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