Given the word $w = ab^p ab^p$, clearly $w\in L \land |w|=2\cdot(p+1)\geq p$.

Now, consider the division $w=xyz$ where $|xy|\leq p \land |y|>0$ then

• Case 1: $x=\varepsilon$, $\forall s\geq0\ ,y=ab^s$ then if $i=0$ we get that $xy^0z$ is actually $b^{p-s}ab^p$ and that's clearly not in L.

• Case 2: $x\not=\varepsilon$ then $\forall s>0:\ y=b^s$ then again if $i=0$ we get that $xy^0z = ab^{p-s}ab^p$ and because $s>0$ then there is not a repeatable word (because $u=ab^p$ or $u=ab^{p-s}$) so we can conclude that $w\notin L$.

Thus $L\notin REG$.

Given the word $w = ab^p ab^p$, clearly $w\in L \land |w|=2\cdot(p+1)\geq p$.

Now, consider the division $w=xyz$ where $|xy|\leq p \land |y|>0$ then

• Case 1: $x=\varepsilon$, $\forall s\geq0\ ,y=ab^s$ then if $i=0$ we get that $xy^0z$ is actually $b^{p-s}ab^p$ and that's clearly not in L, because of the '$a$' in the middle of the new word.

• Case 2: $x\not=\varepsilon$ then $\forall s>0:\ y=b^s$ then again if $i=0$ we get that $xy^0z = ab^{p-s}ab^p$ and because $s>0$ then there is not a repeatable word (because $u=ab^p$ or $u=ab^{p-s}$) so we can conclude that $w\notin L$.

Thus $L\notin REG$.

Given the word $w = ab^p ab^p$, clearly $w\in L \land |w|=2\cdot(p+1)\geq p$.

Now, consider the division $w=xyz$ where $|xy|\leq p \land |y|>0$ then $\forall s\geq0\ ,y=ab^s$ then if $i=0$ we get that $xy^iz$ is actually $b^{p-s}ab^p$ and that's clearly not in L. thus we can conclude that $L\notin REG$.

Given the word $w = ab^p ab^p$, clearly $w\in L \land |w|=2\cdot(p+1)\geq p$.

Now, consider the division $w=xyz$ where $|xy|\leq p \land |y|>0$ then

• Case 1: $x=\varepsilon$, $\forall s\geq0\ ,y=ab^s$ then if $i=0$ we get that $xy^0z$ is actually $b^{p-s}ab^p$ and that's clearly not in L.

• Case 2: $x\not=\varepsilon$ then $\forall s>0:\ y=b^s$ then again if $i=0$ we get that $xy^0z = ab^{p-s}ab^p$ and because $s>0$ then there is not a repeatable word (because $u=ab^p$ or $u=ab^{p-s}$) so we can conclude that $w\notin L$.

Thus $L\notin REG$.