If the axiom $A_{DFA}$ is known to be decidable, would simply constructing a $DFA$ diagram be enough to prove that $L(M)$ is decidable?
Most of what I can find on the internet tells me that you need to construct a decider (IE, a Turing Machine) in order to prove that a language is decidable. But since it's already established that all languages recognizable by $DFA$'s are decidable, it should be adequate to just show the $DFA$, right?
Best Answer
Yes - if you want to show that a language $L$ is decidable, it is enough to construct a DFA $M$ such that the language accepted by $M$ is precisely $L$ - that is, $L(M)=L$.
The set of pairs $(M, w)$ such that $w$ is a word and $M$ is a DFA which accepts $w$ is decidable. In principal, this is slightly stronger than the decidability of each $L(M)$, since this is a statement about uniform decidability.