Trying to break the proof of undecidability of the halting problem
Let's do this step by step.

&quot;Z(A,b) where {A,b!=Z}&quot;
I understand this to mean &quot;just forbid input Z for algorithm Z&quot;. You can not do that. A machine that is supposed to solve the halting problem has to work on all machines, by definition of the problem. In other words, you make the assumption &quot;Let Z be a machine that does not solve the Halting problem&quot; -- that does not lead anywhere.

dkuper's argument is correct: you can derive the contradiction by feeding any of the infinitely many indices of the (assumed) solver of the halting problem, arriving at the same contradiction. Since this set of indices is not recursive, you can not program your machine to detect every alias.

Then the machine does not solve the halting problem anymore (see 1). Even if we had some leeway there, just assume the function that solves the halting problem for all functions but the one which solves the halting problem, and do the same proof.
And when you try to fix this in the same way, I'll get another function. And another one. There are enough of them.


In other words, your strategy of fixing one (or finitely many) special cases won't lead you anywhere. The halting problem is far too undecidable for that -- as the proof shows.