We define the circuit for a select-union-project-join $Q$ recursively by cases as follows. In each case, let $\tuple{V_{Q_i,\pxdb}, E_{Q_i,\pxdb}, \phi_{Q_{i},\pxdb}, \ell_{Q_i,\pxdb}}$ denote the circuit for subquery $Q_i$.
\caseheading{Base Relation}
Let $Q$ be a base relation $R$. We define one node for each tuple. Formally, let $V_{Q,\pxdb}=\comprehension{v_t}{t\in R}$, let $\phi_{Q,\pxdb}(t)= v_t$, let $\ell_{Q,\pxdb}(v_t)= R(t)$, and let $E_{Q,\pxdb}=\emptyset$.
This circuit has $|R|$ vertices.
\caseheading{Selection}
Let $Q =\sigma_\theta\inparen{Q_1}$.
We re-use the circuit for $Q_1$. %, but define a new distinguished node $v_0$ with label $0$ and make it the sink node for all tuples that fail the selection predicate.
Formally, let $V_{Q,\pxdb}= V_{Q_1,\pxdb}$, let $\ell_{Q,\pxdb}(v_0)=0$, and let $\ell_{Q,\pxdb}(v)=\ell_{Q_1,\pxdb}(v)$ for any $v \in V_{Q_1,\pxdb}$. Let $E_{Q,\pxdb}= E_{Q_1,\pxdb}$, and define
$$\phi_{Q,\pxdb}(t)=
\phi_{Q_{1}, \pxdb}(t) \text{ for } t \text{ s.t.}\;\theta(t).$$
Dead sinks are iteratively removed, and so
%\AH{While not explicit, I assume a reviewer would know that the notation above discards tuples/vertices not satisfying the selection predicate.}
%v_0 & \textbf{otherwise}
%\end{cases}$$
this circuit has at most $|V_{Q_1,\pxdb}|$ vertices.
\caseheading{Projection}
Let $Q =\pi_{\vct A}{Q_1}$.
We extend the circuit for ${Q_1}$ with a new set of sum vertices (i.e., vertices with label $+$) for each tuple in $Q$, and connect them to the corresponding sink nodes of the circuit for ${Q_1}$.
Naively, let $V_{Q,\pxdb}= V_{Q_1,\pxdb}\cup\comprehension{v_t}{t \in\pi_{\vct A}{Q_1}}$, let $\phi_{Q,\pxdb}(t)= v_t$, and let $\ell_{Q,\pxdb}(v_t)=+$. Finally let
This formulation will produce vertices with an in-degree greater than two, a problem that we correct by replacing every vertex with an in-degree over two by an equivalent fan-in tree. The resulting structure has at most $|{Q_1}|-1$ new vertices.
% \AH{Is the rightmost operator \emph{supposed} to be a $-$? In the beginning we add $|\pi_{\vct A}{Q_1}|$ vertices.}
The corrected circuit thus has at most $|V_{Q_1,\pxdb}|+|{Q_1}|$ vertices.
\caseheading{Union}
Let $Q ={Q_1}\cup{Q_2}$.
We merge graphs and produce a sum vertex for all tuples in both sides of the union.
Formally, let $V_{Q,\pxdb}= V_{Q_1,\pxdb}\cup V_{Q_2,\pxdb}\cup\comprehension{v_t}{t \in{Q_1}\cap{Q_2}}$, let $\ell_{Q,\pxdb}(v_t)=+$, and let
Let $\ell_{Q,\pxdb}(v_t)=\times$, and let $\phi_{Q,\pxdb}(t)= v_t$
As in projection, newly created vertices will have an in-degree of $k$, and a fan-in tree is required.
There are $|{Q_1}\bowtie\ldots\bowtie{Q_k}|$ such vertices, so the corrected circuit has $|V_{Q_1,\pxdb}|+\ldots+|V_{Q_k,\pxdb}|+(k-1)|{Q_1}\bowtie\ldots\bowtie{Q_k}|$ vertices.
Proof by induction. The base case is a base relation: $Q = R$ and is trivially true since $|V_{R,\pxdb}| = |R|$.
For the inductive step, we assume that we have circuits for subplans $Q_1, \ldots, Q_n$ such that $|V_{Q_i,\pxdb}| \leq(k_i-1)\qruntime{Q_i,\pxdb}$ where $k_i$ is the degree of $Q_i$.
\caseheading{Selection}
Assume that $Q =\sigma_\theta(Q_1)$.
In the circuit for $Q$, $|V_{Q,\pxdb}| = |V_{Q_1,\pxdb}|$ vertices, so from the inductive assumption and $\qruntime{Q,\pxdb}=\qruntime{Q_1,\pxdb}$ by definition, we have $|V_{Q,\pxdb}| \leq(k-1)\qruntime{Q,\pxdb}$.
% \AH{Technically, $\kElem$ is the degree of $\poly_1$, but I guess this is a moot point since one can argue that $\kElem$ is also the degree of $\poly$.}
% OK: Correct
\caseheading{Projection}
Assume that $Q =\pi_{\vct A}(Q_1)$.
The circuit for $Q$ has at most $|V_{Q_1,\pxdb}|+|{Q_1}|$ vertices.
% \AH{The combination of terms above doesn't follow the details for projection above.}