Question $3$: Using the Pumping lemma to prove that the given language is not context free. $[3$ mark$]$ Language:

$\{$a$^$m b$^$n c$^($mn$)$ :m$>=0,$ n$>=0\}\{$a$^$n b a$^(2$n$)$ b a$^(3$n$)$ : n$=>0\}\{$a$^$n b$^$n a$^$n b$^$n :n$>=0\}$ sove it as cases like this:

Case $1$: The substring $vxy$ does not contain any $c.$

Consider the string $u{v}^{2}x{y}^{2}z=$uvvxyyz. Since $|vy|1,$ this string

contains more than $p$ many as or more than $p$ many $b$ s$.$ Since it contains

exactly $p$ many $c,$ it follows that this string is not in the language $A.$ This

is a contradiction because, by the pumping lemma, the string $u{v}^{2}x{y}^{2}z$ is in

A$.$

Case $2$: The substring $vxy$ does not contain any $a.$

Consider the string $u{v}^{2}x{y}^{2}z=$uvvxyyz. Since $|vy|1,$ this string

contains more than $p$ many $b$ s or more than $p$ many cs$.$ Since it contains

exactly $p$ many as$,$ it follows that this string is not in the language $A.$ This

is a contradiction because, by the pumping lemma, the string $u{v}^{2}x{y}^{2}z$ is in

A$.$

Case $3$: The substring vxy contains at least one a and at least one $c.$

Since $s=a^p{b}^p{c}^p,$ this implies that $|vxy|>p,$ which again contradicts the

pumping lemma.

Thus, in all of the three cases, we have obtained a contradiction. There$-$

fore, we have shown that the language $A$ is not context$-$free.