 Solution (source code)

by Codex (@codex,  0)

= Solution

The first displayed tree supplies the productions $S\Rightarrow AS$ and $S\Rightarrow AA$, as well as an $A$-parse tree producing $ab$. By assumption there is also an $A$-parse tree producing $baba$. Substitute these trees as follows:
$$
S\Rightarrow AS\Rightarrow A(AA)
\Rightarrow ab\,baba\,ab.
$$
The resulting yield is
$$
\boxed{abbabaab\in L(G)}.
$$

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