Update on Overleaf.

introduction.tex

@@ -240,9 +240,9 @@ Our negative results (\Cref{tab:lbs}) indicate that \abbrCTIDB{}s (even for $\bo
 We adopt the two-step intensional model of query evaluation used in set-\abbrPDB\xplural, as illustrated in \Cref{fig:two-step}:
 (i) \termStepOne (\abbrStepOne): Given input $\tupset$ and $\query$, output every tuple $\tup$ that possibly satisfies $\query$, annotated with its lineage polynomial $\poly(\vct{X})%=\textcolor{red}{CHANGE}\apolyqdt\inparen{\vct{X}}$
 $;
-(ii) \termStepTwo (\abbrStepTwo): Given $\poly(\vct{X})$ for each tuple, compute an $(1\pm \eps)$-approximation $\expct_{\randWorld\sim\bpd}\pbox{\poly(\vct{\randWorld})}$.
+(ii) \termStepTwo (\abbrStepTwo): Given $\poly(\vct{X})$ for each tuple, compute a $(1\pm \eps)$-approximation $\expct_{\randWorld\sim\bpd}\pbox{\poly(\vct{\randWorld})}$.
 Let $\timeOf{\abbrStepOne}(\query,\tupset,\circuit)$ denote the runtime of \abbrStepOne when it outputs $\circuit$ (a representation of $\poly$ as an arithmetic circuit --- more on this representation in~\Cref{sec:expression-trees}).
-Denote by $\timeOf{\abbrStepTwo}(\circuit, \epsilon)$ (recall $\circuit$ is the output of \abbrStepOne) the runtime of \abbrStepTwo (when $\poly$ is input as $\circuit$). Then
+Denote by $\timeOf{\abbrStepTwo}(\circuit, \epsilon)$ (recall $\circuit$ is the output of \abbrStepOne) the runtime of \abbrStepTwo (when $\poly$ is input as $\circuit$). Then to answer if we can compute 
 %which we can leverage~\Cref{def:reduced-poly} and~\Cref{lem:tidb-reduce-poly} to address the next formal objective:
 
 \begin{Problem}[\abbrCTIDB linear time approximation]\label{prob:big-o-joint-steps}