First Sets: If $A \to Ad\ |\ c$, what is $First(A)$? Suppose that we have a grammar with the following rules:

$$S \to Aa\ |\ b\ |\ \varepsilon\\
A \to Ad\ |\ c$$

From looking at it I already know that $First(S) = \{b, \varepsilon, c\}$. My question is:


How to determine $First(A)$?


For instance, by looking at the rule $S \to Aa$ we know that $First(S)$ must include $First(A)$. $First(A)$ already includes $c$ (that's why $c \in First(S)$). but how to handle the rule $A \to Ad$.

Just by applying the algorithm for finding the $First$ set blindly, I will have that $First(A)$ "includes" $First(A)$, which confuses me.

As far as I understand in $A \to Ad\ |\ c$ there is a left recursion. By eliminating it we get the following rules

$A \to cA'\\
A' \to dA'$

Does this mean that $First(A)$ only contains $c$?

Does this mean that I can ignore a rule with left recursion and just look at the other (non-recursive) rules under that variable?

Thanks!
The grammar you have presented is left linear grammar. The production $A \rightarrow Ad$ in when transformed into $FM$, $A$ becomes state &amp; an edge with $c$ terminal from $A$ to $A$ (self loop). So its clear for this particular grammar $First(A)$ is $c$. 

Moreover, your approach is also correct or primarily correct.