Is this a CFL? $$\{a^n b^m c^k\; : \;k>m \; \; k>n\}$$
When on seeing $a$'s and $b$'s I push them onto stack and as I see $ c$ as input if $ TOS$ is $b$ ,I pop them ,again if $TOS$ is a,I pop them so when my $TOS$ of stack is empty and input is$ c $,I do not make any changes in the stack and that clearly implies that i have no of $c$'s to be greater than both $a$ and $b$.
But its supoosedly not a CFL. Need help where I'm going wrong in constructing the PDA. Because supposedly a PDA can be made for this which makes it a CFL.
With very minor variations your idea can be used to design a PDA that recognizes either of the languages $\{a^nb^mc^k:k\ge n+m\}$ and $\{a^nb^mc^k:k>n+m\}$, so these are both context-free. A similar idea also yields PDAs for $\{a^nb^mc^k:k>n\}$ and $\{a^nb^mc^k:k>m\}$. The problem with the language in your question is that while we can use the stack to keep track of $n$, $m$, or $n+m$, we can’t use it to keep track of both $n$ and $m$.
Let $L=\{a^nb^mc^k:k\ge n\text{ and }k\ge m\}$; we can use the pumping lemma for context-free languages to show that $L$ is not context-free. Suppose that it is, and let $p$ be the pumping length. Let $s=a^pb^pc^p\in L$. The pumping lemma says that we can decompose $s$ as $s=uvxyz$ in such a way that $|vy|\ge 1$, $|vxy|\le p$, and $uv^kxy^kz\in L$ for each $k\ge 0$. Since $|vxy|\le p$, $vxy$ must be a substring either of $a^pb^p$ or of $b^pc^p$. In the former case $uv^kxy^kz\notin L$ whenever $k>1$, since $uv^kxy^kz$ then has more than $p$ $a$s or more than $p$ $b$s but still only $p$ $c$s.
The latter case is a little trickier. If $|y|>0$, take $k=0$: $uxz$ still has $p$ $a$s, but it has fewer than $p$ $c$s, so it’s not in $L$. If $|y|=0$, then $|v|>0$. If $v$ contains at least one $c$, we can again take $k=0$. Otherwise, $v$ consists entirely of $b$s, and $uv^kxy^kz\notin L$ for $k>2$, since $uv^kxy^kz$ then has more than $p$ $b$s but only $p$ $c$s.