I'm using the following definition of $NP$:$$A \in NP \Longleftrightarrow A(x) = \exists w: B(x,w) $$where $B \in P$ and $|w| = poly(|x|)$.  Now instead of the problem whether the program $\Pi$ halts on input $x$, I'll use the close cousin that asks whether $\Pi$ halts on $0$. Then I can write:$$HaltsOnZero(\Pi) = \exists t: HaltsOnZeroInTime(\Pi, t) $$where $HaltsOnZeroInTime$ is the decidable problem of checking whether $\Pi$ halts on input $0$ after $t$ steps. Comparing this with the definition of $NP$ $HaltsOnZeroInTime$ is already in $P$ as long as $t$ is encoded in unary. After all running $\Pi$ for $t$ steps takes only about $t$ steps, so this is a linear algorithm.The only issue is that $t$ isn't in $poly(|\Pi|)$ because programs of size $|\Pi|$ when given $0$ as input run for $BusyBeaver(|\Pi|)$ time, which isn't even computable, much less polynomial. But now suppose we insist that $\Pi$ is padded until it reaches length $BusyBeaver(|\Pi|)$ and call that new input $\Pi'$. Then $t$ would be polynomial (in fact linear) in the size of $\Pi'$ and $HaltsOnZero(\Pi')$ would be in $NP$.At this point there's of course a voice inside me shouting that this is all nonsense because $BusyBeaver$ isn't computable. Well I certainly haven't reduced the classical formulation of the Halting Problem to a problem in $NP$ because I had an uncomputable step in the reduction. But it seems to me like I've still defined a problem in $NP$ with the strange property that just writing down a problem instance requires uncomputable superpowers.Is this really a bona fide $NP$-problem or am I making a subtle mistake somewhere?
Is this padded version of the Halting Problem in NP?
np;halting problem;busy beaver