One of the conditions for A → α | β to be LL(1) grammar states that
If β → ε in one or more steps then α doesn't derive any string beginning with a terminal in FOLLOW(A). Likewise If α → ε in one or more steps then β doesn't derive any string beginning with a terminal in FOLLOW(A).
I couldn't understand what is the intuitive meaning of this condition and why this condition should hold? I could easily understand if an example is provided for the answer.
Considering the above condition, should α derive a string beginning with a terminal in FOLLOW(A), then it becomes impossible to determine which derivation will be used to produce this symbol with lookahead limited to one symbol.
For instance, should $\mathrm{s}$ be the symbol that is both on FIRST(α) and FOLLOW(A), and $x$ the leftmost substring already parsed, then the following sequences are both possible:
$xA$ω $ \Rightarrow x$βω $ \Rightarrow^* x$ω $ \Rightarrow^* x\mathrm{s}$…
$xA$ω $ \Rightarrow x$αω $ \Rightarrow^* x\mathrm{s}$…ω
Even if the grammar is not ambiguous, you can't decide which path to take by reading just $\mathrm{s}$.
