Is the halting problem always decidable for non-universal programs?

For every non-universal computable program $P$ that takes input of type $D$ does there exist some total computable function $g$ that takes an input $I$ of type $D$ and decides successfully whether $P$ halts on $I$?

Some concrete definitions:

A program $P$ that takes some data type $D$ as input is computable iff there exists some total computable function $h$ that takes a input string and returns a $D$, and some Turing machine $T$, which takes a binary string, such that $P(h(I)) \cong T(I)$ for all strings $I$. "$\cong$" includes both not halting as well as both producing the same output.

A program $P$ that takes some data type $D$ as input is universal iff there exists some total computable function $f$ that takes both a description of a Turing Machine and an input string and produces something of type $D$ such that for every Turing machine $T$ and every string $I$, $P(f(T, I)) \cong T(I)$.

To answer the first part of your question - Two counter machines are not exactly Turing-complete. That is, there are some functions that cannot be computed with them, for certain counter-values. This is a very slight weakness, so 2CM are actually very close to being Turing complete, but still.

As for your suggested model - could you formalize it in more detail? It's not clear enough as it is.

StackExchange Computer Science Q#51863, answer score: 5