Is the language $L=\{a^nb^m \mid n>2^m\}$ context-free? I cannot go on with this exercise:


  Determine whether $L = \{a^nb^m \mid n &gt; 2^m \}$ is context-free.


Let's suppose that $L$ is context-free. According to the pumping lemma, there exists $N &gt; 0$ such that every $z \in L$ of size at least $N$ has a decomposition $z = uvwxy$ such that


$|vwx| \leq N$.
$|vx| \geq 1$.
For all $i \geq 0$, $z_i = uv^iwx^iy$ is in $L$.


Let's use $z= a^{2^N+1}b^N$.

Then
$|z| = 2{^N+1} +N &gt; N$ and $v= a^h$ and $x= b^k$ with $1 \leq h+k \leq N$.

So $z_i = a^{2^N+1}a^{h(i-1)} b^{N-k}b^{k(i-1)}$.

So if there exists $i &gt; 0$ such that $2^N+1+h(i-1) \leq 2^{N+(i-1)k}$, then $z_i \notin L$.

How can I go on to show that $L$ is not context-free?
According to Parikh's theorem, if your language were context-free, then the set S = $\{(n,m) : n &gt; 2^m\}$ would be semi-linear. Every linear subset of $S$ has a constant second coordinate, as we show below. Therefore you need an infinite number of linear sets to cover $S$, showing that it is not semi-linear.

It remains to show that every linear subset of $S$ has constant second coordinate. Indeed, consider any linear subset of $S$. By definition, such a subset is of the form $u_0 + \mathbb{N} u_1 + \cdots + \mathbb{N} u_r$. By assumption there exists some $i &gt; 0$ such that $u_i = (a,b)$, where $b \neq 0$. Let $u_0 = (n',m')$. It is not hard to check that for some $t$, $n' + at \leq 2^{m' + tb}$, since the left-hand side grows linearly, whereas the right-hand side grows exponentially. This contradicts the fact that the subset is a subset of $S$.