I need some help with explaining why a grammar is not LL(1).
Let us take the following grammar:
$$ \begin{align} S \rightarrow & aB \mid bA \mid \varepsilon \\ A \rightarrow & aS \mid bAA \\ B \rightarrow & b \\ \end{align} $$
This is my attempt:
For the grammar to be LL(1) it is a necessary condition that for any strings $c_1γ$ and $c_2β$, derivable from $S \rightarrow aB$ and $A \rightarrow aS$ respectively, we have $c_1 \ne c_2$.
But, $S \rightarrow aB$ and $A \rightarrow aS$, hence $c_1 = c_2$ and the grammar is not LL(1).
Is my reasoning right?
Thanks in advance.