I've read that it is known that tag systems in TS(2,2), that is, with two symbols and deletion number $2$, are decidable.
I suppose I'm asking for clarification on what that means in this context. I assume it means that at the very least, for any given system in that class, one can effectively calculate whether or not it eventually halts.
But what about more specific questions? In particular, I believe the simplest non-trivial system in TS(2,2) is $$\{a \rightarrow aab, b\rightarrow a\}.$$
It is immediately clear that almost all starting values grow without bound, and beyond that, there are definite patterns you can make headway on. However, it is not at all clear to me that, say, there's any closed-form way to determine whether the $k$th symbol added will be an $a$ or a $b$, or what the parity is after some number of rounds, or anything like that. My understanding is that Rice's Theorem wouldn't apply since a decidable system can't be Turing complete, so that doesn't answer it. So we know that any result we want can only take finitely long to calculate, but that leaves my main question:
Is it possible for a computational system to be decidable, yet have computationally irreducible properties which have no shortcuts and thus require actually doing all the intervening computation to determine?
It looks like it specifically says it’s not decidable and is Turing complete.
Also, apparently the particular case for having an alphabet with just two letters is not an exception, unless I am misunderstanding something else.
Where did you read it was decidable?