# Tag Info

Having a verifier for a language in general is known as semidecidability, which is actually weaker than decidability. So in general, the answer is no, we can't build a decider for $L$. But if the verifier is efficient (i.e., $L \in$ NP), then indeed $L$ is decidable, and your argument is pretty close for why that is. Let's say we have an efficient verifier \$...