\]

Ideally we would now like to sum this expression over all possible paths $\xi_1,\xi_2$ and use $p_\mathrm{tot}:=\sum_{\xi\in\paths{b_i}} \mathbb{P}[\xi] = 1$ (which also holds up to arbitrary order in $p$). The above expression would then become $R_{b_1} + R_{b_2}$. However, not all paths in the sum would satisfy the independence condition so it seems we can't do this. We now argue that it works up to order $p^{k-1}$.

For all $\xi\in\paths{b_1\land b_2}$ we have that \emph{either} $\xi$ splits into two independent paths $\xi_1,\xi_2$ as above, \emph{or} it does not. In the latter case, when $\xi$ can not be split like that, we know $\mathbb{P}[\xi]$ contains a power $p^k$ or higher because there is a gap of size $k$  and the paths must have moved at least $k$ times `towards each other' (for example one path moves $m$ times to the right and the other path moves $k-m$ times to the left). So the total weight of such a combined path is at least order $p^k$. Therefore we have

\[

	R_{b_1\land b_2} = \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_1]\mathbb{P}[\xi_2]|\xi_2| + \sum_{\mathclap{\xi\;\mathrm{dependent}}} \mathbb{P}[\xi]|\xi|.

\]

\[

	\sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    = \sum_{\xi_1\in\paths{b_1}} \sum_{\substack{\xi_2\in\paths{b_2}\\ \text{independent of }\xi_1}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

\]

\end{proof}

\begin{center}

\includegraphics{diagram_paths.pdf}

\end{center}

\textbf{Proof of claim \ref{claim:weakcancel}}: Say we have a group on the left with $l$ slots and a group on the right with $r$ slots, with enough space between the groups. Then on the left we have strings in $\{0,1'\}^l$ as possibilities and on the right we have strings in $\{0,1'\}^r$. The combined configuration can be described by strings $(a,b)\in\{0,1'\}^{l+r}$. Such a configuration has probability $(-1)^{|a|+|b|} p^{r+l}$ in $\rho$ and by claim \ref{claim:expectationsum} we know $R_{(a,b)} = R_a + R_b + \mathcal{O}(p^\mathrm{spacing})$. The total contribution of these configurations is therefore

\begin{align*}

    To actually go below $2k$ one needs to be careful, because there are periodic configurations that are invariant under some rotations causing double counting issues. This can be probably resolved by showing that when a pattern becomes periodic for some $n$ it actually produces periodicity times more expectation due to symmetry. But this is all just speculation.

\end{comment}

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