Suppose we have Turing machine $M^*$ that:
i. halts printing 1 if $M_n$ halts on input 1
ii. halts printing 0 if $M_n$ doesn't halt on input 1
Show that you cannot construct $M^*$.
Suppose we run $M^*$ on itself so that:
i. $M^*$ halts printing 1 if $M^*$ halts on input 1
ii. $M^*$ halts printing 0 if $M^*$ doesn't halt on input 1
ii. is a contradiction, so it's impossible to construct $M^*$.
I'm not sure if this "proof" works or how I can proceed to show that $M^*$ is impossible. I know there is a way to show that it leads to making the halting problem decidable, but I don't know how to show that.