I need some help to prove that the language is decidable.
$K$ = {$N$ : $N$ is a DFA (Sigma = {a, b, c}) and $L$($N$) contains at least one word in which there is no a}.
It tried to make an algorithm which receives as input a DFA $N$ and which determines whether or not this automaton accepts at least one word which contains no a. But it's kinda complex so any help would be appreciated.
A DFA is essentially a directed multi-graph $G$ in which a distinguished vertex is marked as the initial state $s$, a set $T$ of vertices are marked as accepting states, and edges are labelled with the symbols of $\Sigma$.
The problem is then equivalent to that of deciding whether there is a path on $G$ that goes from $s$ to a vertex in $T$, and uses only edges that are not labelled with $a$.
It suffices to delete all the edges labelled with $a$ from $G$ and to run any graph visit algorithm from $s$. If a vertex in $T$ is discovered you can immediately accept. If the visit terminates and no vertex in $T$ is discovered, then reject.
