So the Halting Problem states (or at least one statement is) that there isn't a Turing Machine that decides whether or not a given Turing Machine halts on a given input, using only its description.
I saw a proof of it in Richard Lassaigne's book, Logic and Complexity, which uses a diagonalization arguement. It proves that the problem of checking whether a given Turing Machine halts, when the input is its own description, is undecidable. From here we can conclude that the Halting Problem is undecidable as well, since it "contains" the other problem.
My question is about the part that's left out, so checking whether a TM halts on a given input, excluding its own description. Is there a simple way to show that this is also an undecidable problem?
Since the diagonalization depends on the enumeration you take, I thought about getting a new enumeration, that would compute, for example, the TM with description $m$, when given the number $m+1$. Then I could have the same arguement as for the Halting Problem, but for the diagonal above the main one and all the other "diagonals".
Is this a valid arguement?
For any machine $M$ on any input $x$, you can (using snm) prove that there is a machine $M_x$ that do the same compution than $M$ on input $x$, on any input :
$$\forall M\forall x \exists M_x\forall y \quad M(x)=M_x(y)$$
Then, all the information is in the diagonal, because $M_x$ on its own description compute also $M(x)$.