Use the pumping lemma to show that the language is not context-free I want to show that this language is not context-free:

I struggle with the pumping lemma for CFL and i would like to know how to solve this problem. Thank you
(skipping the classic introductory text of the proof)
During my proof i chose the word $z = a^pb^pc^{p-1}$ ( correct me if this isn't a good choice)
The word z can be cut as $sv^iuw^it$
Let's take i = 2
From there i know that we have to explore all cases:
Case 1: v and w are only made of a's so we have $a^{p+|vw|}b^pc^{p-1}$ so this isn't in the language because $|vw| >= 1$
Case 2: it's the same thinking but v w are only made of b's
Case 3: v and w are only made of c's, if $i=2$ then we have $v^2$and $w^2$ so $|vw|=4$ so we have $a^pb^pc^{p-1+|vw|}$ = $a^pb^pc^{p-1+4}$ which mean here that it won't be in the language since we need $m>=n$
From there ( if everything yet makes sens) i am not too sure on how to continue this problem, could anyone help out ? Thanks a lot.
 A: The tricky cases are when there is an $r\le\frac{p}2$ such that $v=a^r$ and $w=b^r$, because then if you make $i>1$, $sv^iuw^it$ is still in $L$. However, the pumping lemma says that $sv^iuw^it\in L$ for all $i\ge 0$, so you can pump down to $i=0$, so that $sv^iuw^it=a^{p-r}b^{p-r}c^{p-1}$. If $r>1$, this word is not in $L$, because $p-r<p-1$. Unfortunately, if $r=1$ the word is $a^{p-1}b^{p-1}c^{p-1}$, which is in $L$. The problem here is with your choice of $z$: you should have let $z=a^pb^pc^p$. This is in $L$, since the definition of $L$ requires only that $m\ge n$, not that $m>n$, and now everything works, both the cases that you’d already considered and the one that I just did.
The remaining possibilities are:

*

*$v=b^r$ for some $r\ge 1$ and $w=c^s$ for some $s\ge 1$, in which case pumping up or down will give you unequal numbers of $a$s and $b$s and hence a word not in $L$;

*$v=a^rb^s$ for some $r,s\ge 1$, in which case $w$ cannot contain any $c$s (because $|vuw|\le p$), so pumping down will drop the number of $a$s below the number of $c$s and give you a word not in $L$; and

*$v=b^rc^s$ for some $r,s\ge 1$, in which case $w$ contains only $c$s, and pumping up or down gives you unequal numbers of $a$s and $b$s and hence a word not in $L$.

In every case, therefore, we can pump $z$ to get a word not in $L$, and the pumping lemma then implies that $L$ is not context free.
A: Choose the string w =  a^p b^p c^p  clearly w is in L , we provide an easy way to show that w cannot be pumped  ( we cut the string into u v^i x y^i z)

*

*either v or y has more than one type of symbol :

If either v or y has more than one type of symbol , then when pumping w , the string w' = u v^2 x y^2 z will not have symbols in correct order , so v and y each must contain only one type of symbols ( ex v has only as and y has only bs , or both have only as , etc )


*either v or y has cs :

Assume y has cs , from 1 we know y can have only cs , now v also can have one type of symbols , consider the string      w' = u v^2 x y^2 z , if we choose v to be also cs , then w' has more cs than as and bs , if we choose v to be bs , w' has more bs than as , if we choose v to be as ( which we cannot since |vxy| > p , but anyway ) we have more as than bs , so v and y cannot have cs
3 ) final case
From 2 we know that v and y cannot have cs , and from 1 we know each must have one type of symbols , if both v and y have only as or only bs , then w' = u v^2 x y^2 z would have more as than bs , or more bs than as , if v has only as and y has only bs , then , w'' = u v^0 x y^0 z ( a very useful case which is usually forgotten)  , would have more cs than as and bs
Conclusion :

*

*tells us v and y each must contain only one type of symbols


*tells us v and y cannot contain cs


*tells us any way we choose v and y to be as or bs fails
So w cannot be pumped and L is not a CFL
