I was reading the following set of notes on logic (page 84 of paper pdf) and came a across a rather simple definition but was unsure what it meant conceptually:
I think I understand the proof (it would be more clear if the characteristic function was involved) but it didn't feel like I understood what the definition actually meant conceptually or why it was intereting.
It seems that whenever one of the condition $R_i(a)$ is true then we output whatever $P_i(a)$ does. $R(a) \iff a \in R \subseteq \mathbb N^n$. So I am assuming that $P(a)$ is just outputting the relation it is suppose to define. Informally $P:\mathbb N^n \to \{0,1\}$ (though earlier they actually define the characteristic function for this, i.e. indicator function for this). So to me it just seems rather odd to do this because we are defining a new relation using an old one. So Basically whenever we have that the old relation hold we actually return a different relation (or indicator of it). Which seems really weird to me. It's nearly like the new relation function $P$ is a liar. It actually uses some hidden relation $R$ to compute itself but returns other elements. It nearly feels that $R_i$ is irrelevant because if I saw the specification of $P$ (say as a programer), I would just see what it really computes, which is $P_i$. If it does some weird thing in the background seems rather irrelevant.
I suspect I either have a misconception or misinterpreting things (or over thinking things?), though the definition for this new relation $P$ seems poorly motivated to me because the assumption is that $P_i$ is computable, so why do we even need $R_i$ to compute $P_i$? I just don't see the point.
PS: didn't know what would be a good title for this...
I'd write $P = (P_1\cdot R_1) + \ldots + (P_n\cdot R_n)$. Then for each input $a$, $$P(a) = (P_1(a)\cdot R_1(a)) + \ldots + (P_n(a)\cdot R_n(a)).$$ If $R_i(a)=1$ and $R_j(a)=0$ for each $j\ne i$, then $P(a) = P_i(a)\cdot R_i(a) = P_i(a)$ as claimed. If $P_1,\ldots,P_n$ and $R_1,\ldots,R_n$ are computable, then $P$ is computable.