Not respecting the 3rd rule of a LL(1) grammar

I'm trying to build a grammar that violate only the 3rd rule. I'm trying to figure out what kind of grammar would not respect that.

I think the grammar has to be left-recursive to not respect it.

if $\beta \Rightarrow^* \epsilon$ then $\alpha$ does not derive any string beginning with a terminal in $\mathop {FOLLOW}(A)$. Where $A \to \alpha \mid \beta$

Am I right to think that?

• What "rules" are you talking about? Presumably it is enough to just add a production (or two) that violate the rule. Feb 18 '16 at 23:34
• Added the rule. Feb 18 '16 at 23:47

\begin{align} S &\to A a \\ A &\to a \mid \epsilon \end{align}
• ok so FOLLOW(A) = {a,$} in this case. And FIRST(S) = {a} Feb 19 '16 at 0:48 • @Laura, what is important is that$A \Rightarrow^* \epsilon$,$\mathop{FIRST}(A) = \{a\}$and$\mathop{FOLLOW}(A) = \{a\}$, so$\mathop{FIRST}(A) \cap \mathop{FOLLOW}(A) \ne \varnothing\$ Feb 19 '16 at 1:21