Let $f(x,t)=\begin{cases} 1 \quad \text{if Prime(x) is true }\\ 0 \quad \text{otherwise}\end{cases}$
Clearly f(x,t) is a computable predicate, since Prime(x) is primitive recursive and definition by cases is also primitive recursive.
Could we say $\{x|(\forall t)f(t,x)\}$ is recursively enumerable(r.e). It is the domain of a partially computable function. Here by partially computable function, I mean a program which computes the function which may or may not halt.
I find whenever a predicate f is computable, we could always say $\{x|(\forall t)f(t,x)\}$ is r.e. Is this true?
Could someone give a counterexample of a predicate where this is not true?
How to argue that a set is recursive or recursively enumerable?
Definition of a r.e. http://prnt.sc/9lq22b
Let $A$ be a recursively enumerable set which is not recursive. The complement $\mathbb N\setminus A$ is not recursively enumerable, since a recursively enumerable set with recursively enumerable complement is recursive. Since $A$ is recursively enumerable and nonempty, there is a computable surjection $g:\mathbb N\to A.$ Define a predicate $$f(t,x)=\begin{cases} 1\text{ if }g(t)\ne x,\\ 0\text{ if }g(t)=x. \end{cases}$$ Then $f(t,x)$ is computable, and $\{x|(\forall t)f(t,x)=1\}=\{x|(\forall t)g(t)\ne x\}=\mathbb N\setminus A$ is not recursively enumerable.