I know decidable languages are closed under concatenation and union. But i am having a hard time finding an counter argument (i think its wrong).
Can you help me find the right direction to proof/disproof this lemma ?
2026-02-23 10:14:12.1771841652
$L^*$ decidable $\implies$ $L$ is decidable
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Unfortunately, it isn't true. Let $L$ be the language consisting of the symbols $a$ and $b$ together with an undecidable set of strings in that alphabet. In this case, $L^*$ has all strings, and so it is decidable, but $L$ is not.