I am trying to learn the Rice–Shapiro theorem on my own. Here is the theorem:
Let $Γ$ be a set of computable functions such that the set $R_Γ$ is recursively enumerable. We have $f ∈ Γ$ if and only if there exists a finite function $θ ∈ Γ$ such that $θ ⊆ f$.
I will appreciate if anybody helps me to make the theorem crystal clear. I just want to know what it really means and how I can use it to say that a set is not r.e., for example the set Empty or TOT.
Rice-Shapiro greatly helps in proving that a set is not RE, for some sets of the form $R_\Gamma$.
E.g. take $A = \{ n \ |\ {\sf dom}(\phi_n) \mbox{ finite} \}$. We have $A=R_\Gamma$ with $\Gamma = \{ \phi_n | {\sf dom}(\phi_n) \mbox{ finite} \}$. If $A$ were RE, since e.g. the always undefined function belongs to $\Gamma$, then by Rice-Shapiro any recursive extension of that would be long to $\Gamma$ as well -- but this is false, since e.g. the identity function $\notin\Gamma$. Hence $A$ is not RE.
The above one is a typical use, where you reach a contradiction by extending a function inside $\Gamma$ to one recursive function outside.
We can also use it in the other direction: we reach a contradiction if we can take any function inside $\Gamma$ whose finite restrictions are all outside $\Gamma$. For instance, take $B = \{n \ |\ \forall k\in\mathbb{N}.\ \phi_n(2k)=5 \}$. Clearly $B=R_\Gamma$ with $\Gamma = \{\phi_n \ |\ \forall k\in\mathbb{N}.\ \phi_n(2k)=5 \}$. The function $f(x)=5$ is inside $\Gamma$, but any finite restriction of $f$ can not evaluate to $5$ on all even numbers, hence it must be found outside $\Gamma$. If $B$ were RE, by Rice-Shapiro one such restriction should instead be found inside $\Gamma$. We conclude that $B$ can not be RE.
I find Rice-Shapiro easier to use if I think about it as two separate results:
One is a monotonicity result, telling us that if $f \in \Gamma$ and $f \subseteq g$ with $g$ partial recursive, then $g \in \Gamma$. I.e. recursive extensions of functions in $\Gamma$ must belong to $\Gamma$ -- otherwise, $R_\Gamma$ is not RE.
The other is a compactness result. For any function in $\Gamma$, there must be a finite restriction of it inside $\Gamma$ -- otherwise, $R_\Gamma$ is not RE.
Many, many common examples of non-RE sets violate one of these two conditions, making it simple to prove they are not RE. For instance, consider $C = \{n\ |\ \phi_n(3)\neq 4 \land \phi_n(5)=2 \}$, where the notation $\phi_n(3)\neq 4$ is meant to be true when $\phi_n(3)$ is undefined. This immediately violates monotonicity, since I can always extend a (recursive) function which is undefined on $3$ so that it evaluates to $4$ on that point (and still get a recursive function). So $C$ is not RE.
The theorem says that any observation made about computable functions can be done by testing the values of the functions at finitely many arguments. In particular, acecss to the source code does not give you the ability to make more observations than what you get by trying out what the function does.
