Yo! This is probably a stupid question, however I've never seen it written down explicitly if, for instance, decidability of type-checking is equivalent to the strong normalization property. Therefore I'm asking this question to clarify all possible relations between type-checking, typability and strong normalization.
Let me explain my motivation. For type theories (I'm being intentionally vague here, but I'm interested mainly in dependent type theories), strong normalization is used to prove the decidability of type-checking. On the other side, any typed systems that I know that have one of these properties also have the other one. However I've never seen explicitly stated that strong normalization is equivalent to decidability of type-checking.
Analogously, to prove typability, one usually (maybe always), reduces a term to a normal form. However it's know that typability is not true for dependent type theories, whereas strong normalization may hold.
By decidability of type-checking, I mean that for any given type $A$, context $\Gamma$ and untyped term $a$, it's possible to decide in a finite number of steps whether $\Gamma \vdash a: A$ is true or not.
By decidability of typability, I mean that for any given untyped term $a$, it's possible to decide in a finite number of steps whether there exist a context $\Gamma$ and a type $A$ such that $\Gamma \vdash a: A$ is true.
1)Is it true that decidability of type-checking is equivalent to every term being strongly normalizable?
2)More generally, what's the relation between decidability of type-checking, typability and strong normalization? Which one implies the other?
Thanks in advance.
Given the unsatisfaction regarding the level of generality of my question (which I was unaware of), I would like to delimit it only to Pure Type Systems. Of course, additional comments or counterexamples regarding other type theories will be of great utility.