How to prove $\{ a^ma^lcb^{m+l}\mid m,l \in N \}$ is not regular using the Pumping Lemma?
First of all I would like to write  $M= \left\{ { a }^{ m }{ a }^{ l }c{ b }^{ m+l }\mid m,l \in N \right\} $ as $ M = \left\{ { a }^{ 2n }c{ b }^{ 2n }\mid n \in  N \right\} $. Is this possible?


We have $a^ma^\ell cb^{m+\ell}=a^{m+\ell}cb^{m+\ell}$ so you can certainly write that as $a^ncb^n$ for some $n\geq 2$. But you can't write it as $a^{2n}cb^{2n}$ because $m+\ell$ isn't necessarily even.

The question originally contained a typo, as follows. I've kept my answer to that version of the question because it illustrates a common mistake, so will probably be useful to somebody.


  First of all I would like to write  $M= \left\{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }\mid m,l \in N \right\} $ as $ M = \left\{ { a }^{ 2n }c{ b }^{ 2n }\mid n \in  N \right\} $. Is this possible?


No, you can't do that. Let $M=\{a^ma^\ell cb^{m+1}\mid m,\ell\in\mathbb{N}\}$ and $M'= \{a^{2n}cb^{2n}\mid n\in\mathbb{N}\}$. Any $a^{2n}cb^{2n}\in M'$ can be rewritten as $a^{(2n-1)}a^1cb^{(2n-1)+1}\in M$ (i.e.,  we can set $m=2n-1$ $\ell=1$), so $M'\subseteq M$.  However, take $m=2$, $\ell=1$ and we get $a^3cb^3\in M$, but this string is not in&nbsp;$M'$, which means that $M'\neq M$.

You've tried to prove that $M'$ is non-regular but that doesn't prove that $M$&nbsp;is non-regular. For example, pick your favourite non-regular language&nbsp;$L$ over your favourite alphabet&nbsp;$\Sigma$. $L\subsetneq\Sigma^*$ but the fact that $L$&nbsp;is non-regular certainly doesn't mean that $\Sigma^*$&nbsp;is non-regular!


  Proof.


Your proof that $M'$ is non-regular is basically correct except for two small mistakes. First, you can't assume that $x$ and&nbsp;$y$ have even length: they need to have length $i$ and&nbsp;$j$, rather than $2i$ and&nbsp;$2j$. Remember that you need to show that every partition of the string as $xyz$ fails the pumping property, not just the ones where $x$ and&nbsp;$y$ have some particular property. Second, in your sentences about $xy^0z$, some of the $i$s should be $j$s.