# Is this Language decidable?

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

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

• Turing machines are not decidable (or undecidable). Languages are decidable. – Yuval Filmus Aug 30 at 7:11
• @YuvalFilmus: the title doesn't ask Is this Turing machine decidable?, but Is this Turing-Machine decidable?, and the pixel raster shows a language.) – greybeard Aug 30 at 7:14
• thanks for noting. I'm a bit new to these definitions. – CompuPhisics Aug 30 at 7:48
• It's not decidable. You could use Rice's Theorem to prove that. – plshelp Aug 30 at 12:22

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 $$M'$$.