Why are decidable languages closed under complement?
So if L is decidable why is the complement of L also decidable.
2026-04-02 13:40:57.1775137257
Decidable language closed under complement
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(To get this off the unanswered list, I’m converting my comment to an answer.)
If you have a decision mechanism that always correctly returns yes or no to the question
interchanging the outputs — i.e., turning yes to no and no to yes — gives you a decision mechanism for the complement of $L$.