Confusion in the added loop of the Halting Problem I know there's like a thousand questions about this topic in the site and elsewhere. I'm just going to pick one that at least for me it serves as a good basis for my question. The answer by Rick Decker is the one that I understood more, so I'm just going to take it as a basis for my actual doubt.
It's not difficult to see that if one adds an if (and extra loop) into 'Halting Program '  $H$, then feeding $H$ to itself yields a contradiction. What it confuses me is why to add the extra loop to force the contradiction, if by hypothesis $H$ solves (itself?)
It doesn't seem logical to say first (of course I'm missing something), hey I give you $H$ that solves your issue, but afterwards I give you an augmented $H$ that does not and so we reach a contradiction.
What we show is that this augmented $H$ can not even exist. You gave me $H$, and I gave you back an augmented $H$, called $\hat H$. We can build this $\hat H$ by adding the extra loop and if statement, but its easy to show that this augmented $\hat H$ will never be able to exist: $\hat H(\langle \hat H\rangle)$ is not either rejecting or accepting, nor stuck in an infinite loop, but we know that any TM must be either one of those.
Therefore, we get a contradiction to the existence of $H$: we showed that if $H$ exists, then some other $\hat H$ exists, but also $\hat H$ couldn't ever exist.