# 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.
• Thanks for the explanation of the ancillary $\hat{H}$ that it is what actually does the job. Do you know if it's there a possibility to show this without contradiction? May 11 at 13:46