Existence of particular/specialized halting oracles It is known that there does not exists an oracle $H$ which given any pair $(M,I)$ where $M$ is a machine and $I$ is an input (possibly still a machine) to have $H(M(I)) = YES$ if $M(I)$ halts and $H(M(I)) = NO$ if $M(I)$ does not halt. This was proven by Turing. However, this can be shortly written as
$$ \not\exists H \hspace{0.5cm} \forall M,I \hspace{0.5cm} H(M(I)) = \begin{cases} YES, \hspace{0.3cm}M(I)  \text{  halts }\\ NO, \hspace{0.4cm} M(I) \text{ doesn not halt}\end{cases}$$
But is it possible to have the following:
$$ \forall M,I \hspace{0.5cm} \exists H_{M,I} \hspace{0.5cm} H_{M,I}(M(I)) = \begin{cases} YES, \hspace{0.3cm}M(I)  \text{  halts }\\ NO, \hspace{0.4cm} M(I) \text{ doesn not halt}\end{cases}$$
Yes. Given a machine $M$ and input $I$, one of the two things happens:

$M(I)$ halts.
$M(I)$ does not halt.

We may not be able to predict or compute which of these happens, but one and only one happens simply by rules of classical logic.
Now to show that $H_{M,I}$ exists we consider two cases:

If $M(I)$ halts then $H_{M,I}$ exists because we can take it to be a machine that immediately halts and outputs &quot;yes&quot;.
If $M(I)$ does not halt then $H_{M,I}$ exists because we can take it to be a machine that immediately halts and outputs &quot;no&quot;.

Note: we proved existence of $H_{M,I}$, but we did not give a way of computing $H_{M,I}$ from $M$ and $I$, which is impossible as that would give us the halting oracle (exercise).
So, what Yuval said.