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%\subsection{Reduced Polynomials and Equivalences}

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\AH{\Cref{def:reduced-poly} replaces this.}
\begin{Definition}[Reduced \bi Polynomials]\label{def:reduced-bi-poly}
  Let $\poly(\vct{X})$ be a \bi-lineage polynomial.
  The reduced form $\rpoly(\vct{X})$ of $\poly(\vct{X})$ is the same as \Cref{def:reduced-poly} with the added constraint that all monomials with variables $X_{\block, i}, X_{\block, j}, i\neq j$ from the same block $\block$ are omitted.
\end{Definition}
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%

Consider a $\abbrBIDB$ polynomial $\poly\inparen{\vct{X}} = X_{1, 1}X_{1, 2} + X_{1, 2}X_{2, 1}^2$.  Then by \Cref{def:reduced-bi-poly}, we have that $\rpoly\inparen{\vct{X}} = X_{1, 2}X_{2, 1}$.  Next, we show why the reduced form is useful for our purposes.
%%Removing this example to save space
\iffalse
\begin{Example}\label{example:qtilde}
Consider $\poly(X, Y) = (X + Y)(X + Y)$ where $X$ and $Y$ are from different blocks.  The expanded derivation for $\rpoly(X, Y)$ is
\begin{align*}
(&X^2 + 2XY + Y^2 \mod X^2 - X) \mod Y^2 - Y\\
= ~&X + 2XY + Y^2 \mod Y^2 - Y\\
= ~& X + 2XY + Y
\end{align*}
\end{Example}
\fi
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\AH{Reduction \emph{I think} combined with~\Cref{lem:tidb-reduce-poly} should replace this.  In fact~\Cref{lem:tidb-reduce-poly} works for also for \abbrBIDB\xplural, so, maybe still stating this, but that it follows from~\Cref{lem:tidb-reduce-poly}.}
\begin{Lemma}\label{lem:exp-poly-rpoly}
Let $\pdb$ be a \abbrBIDB over $\numvar$ input tuples such that the probability distribution $\pdassign$ over $\{0,1\}^\numvar$ (the all worlds set) is induced by the probability vector $\probAllTup = (\prob_1, \ldots, \prob_\numvar)$.  As in \Cref{lem:tidb-reduce-poly} for \abbrTIDB, any \abbrBIDB-lineage polynomial $\poly(\vct{X})$ based on $\pdb$ and query $\query$ we have:
  % The expectation over possible worlds in $\poly(\vct{X})$ is equal to $\rpoly(\prob_1,\ldots, \prob_\numvar)$.
\begin{equation*}
\expct_{\vct{W}\sim \pdassign}\pbox{\poly(\vct{W})}  = \rpoly(\probAllTup).
\end{equation*}
\end{Lemma}
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Let $\abs{\poly}$ be the number of operators in $\poly$.

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\begin{Corollary}\label{cor:expct-sop}
If $\poly$ is a \bi-lineage polynomial already in \abbrSMB, then the expectation of $\poly$, i.e., $\expct\pbox{\poly} = \rpoly\left(\prob_1,\ldots, \prob_\numvar\right)$ can be computed in $\bigO{\abs{\poly}}$ time.
\end{Corollary}


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