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As the title says; is this language decidable and how do you prove it?

$$L =\{\langle M\rangle \mid M \text{ is a Turing Machine and there is an input that } M \text{ halts on} \} $$

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  • $\begingroup$ Turing machines are not decidable (or undecidable). Languages are decidable. $\endgroup$ – Yuval Filmus Aug 30 at 7:11
  • $\begingroup$ @YuvalFilmus: the title doesn't ask Is this Turing machine decidable?, but Is this Turing-Machine decidable?, and the pixel raster shows a language.) $\endgroup$ – greybeard Aug 30 at 7:14
  • $\begingroup$ thanks for noting. I'm a bit new to these definitions. $\endgroup$ – CompuPhisics Aug 30 at 7:48
  • $\begingroup$ It's not decidable. You could use Rice's Theorem to prove that. $\endgroup$ – plshelp Aug 30 at 12:22
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Your language is not decidable. To show that it suffices to notice that the existence of a Turing machine $T$ that decided $L$ implies the existence of a Turing machine that solves the halting problem.

Indeed, given any Turing machine $M$ with input $x$ you can construct a Turing machine $M'$ that ignores its input, writes $x$ on the tape, and then simulates $M$. Clearly $M'$ halts (regardless of its input) if and only if $M$ halts on input $x$. We can decide whether $M'$ halts by checking whether $M' \in L$ using $T$ with input $M'$.

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