Are decidable set/languages EQUIVALENT to type 1 grammars (non-contracting)?
Yuval's answer is absolutely right, but note that there's a more general principle in play here as well: diagonalization. Any time we have a class of decidable languages for which we have a &quot;decidable syntactic description&quot; (I'm being vague about this for a moment), we can diagonalize out of this class to get a decidable language not in the class. Basically, we whip up a notion of &quot;$n$th language in the class&quot; and then consider the new language which contains word $1...1$ ($n$ times) iff the $n$th language in the class does not contain that word (or something similar); granting the complexity assumption on the starting class, this results in a decidable language.
(In snappier computability-theoretic terminology: if $f(x,y)$ is a total computable function then there is a total computable $g(z)$ such that for each $x$ there is some $y$ with $f(x,y)\not=g(y)$.)
Classes like type $1$ grammars are defined by simple syntactic conditions, and so can be diagonalized out of. Of course this does not necessarily result in a natural example of a decidable language not in the starting class, but it does prove that that class misses many decidable languages.