Are these two context free grammars equivalent? Let $\Sigma = \{a,b\}$. A CFG for the language $\{a^nb^m~|~n > 2m\}$ can be written as:
$$
\begin{align}
S &\to aaSb\\
S &\to A\\
A &\to aA\\
A &\to a
\end{align}
$$
Would it be equivalent to write this CFG as the following?
$$
\begin{align}
S\to aaSb~|~aS~|~a
\end{align}
$$
Thanks

Yes. They will be the same. 


S-->aaSb,S-->A,
    A-->aA,
    A-->a


and 


S--> S-->aaSb | aS | a


generate the same grammar, .$a^{2n+k}b^n, n \ge 1, k \ge 1$
