%root: main.tex
The evaluation of $\abs{\circuit}(1,\ldots, 1)$ can be defined recursively, as follows (where $\circuit_\linput$ and $\circuit_\rinput$ are the `left' and `right' inputs of $\circuit$ if they exist):

{\small
\begin{align}
\label{eq:T-all-ones}
\abs{\circuit}(1,\ldots, 1) = \begin{cases}
						\abs{\circuit_\linput}(1,\ldots, 1) \cdot \abs{\circuit_\rinput}(1,\ldots, 1)	&\textbf{if }\circuit.\type = \revision{\circmult}\\
						\abs{\circuit_\linput}(1,\ldots, 1) + \abs{\circuit_\rinput}(1,\ldots, 1)		&\textbf{if }\circuit.\type = \revision{\circplus} \\
						 |\circuit.\val|											&\textbf{if }\circuit.\type = \tnum\\
						1													&\textbf{if }\circuit.\type = \var.
					\end{cases}
\end{align}
}

It turns out that for proof of~\Cref{lem:sample}, we need to argue that when $\circuit.\type = +$, we indeed have
\begin{align}
\label{eq:T-weights}
\circuit.\lwght &\gets \frac{\abs{\circuit_\linput}(1,\ldots, 1)}{\abs{\circuit_\linput}(1,\ldots, 1) + \abs{\circuit_\rinput}(1,\ldots, 1)};\\
\circuit.\rwght &\gets \frac{\abs{\circuit_\rinput}(1,\ldots, 1)}{\abs{\circuit_\linput}(1,\ldots, 1)+ \abs{\circuit_\rinput}(1,\ldots, 1)}
\end{align}