Changeset - 53e0f54cd945
[Not reviewed]
0 1 2
Tom Bannink - 8 years ago 2017-06-13 16:49:19
tom.bannink@cwi.nl
Add diagrams and proof of 'last part'
3 files changed with 129 insertions and 10 deletions:
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diagram_patches.pdf
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diagram_patches.tex
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new file 100644
 
\documentclass{standalone}
 
\usepackage[T1]{fontenc}
 
\usepackage{amsmath}
 
\usepackage{amsfonts}
 
\usepackage{parskip}
 
\usepackage{marvosym} %Lightning symbol
 
\usepackage[usenames,dvipsnames]{color}
 
\usepackage[hidelinks]{hyperref}
 
\renewcommand*{\familydefault}{\sfdefault}
 

	
 
\usepackage{bbm} %For \mathbbm{1}
 
%\usepackage{bbold}
 
\usepackage{tikz}
 

	
 
\begin{document}
 

	
 
\begin{tikzpicture}
 
    % Cirlce
 
    \draw[gray] (0,0) -- (10,0);
 
    \draw[dotted] (0,2) -- (10,2);
 
    \draw[gray] (0,2) arc (90:270:1);
 
    \draw[gray] (10,0) arc (-90:90:1);
 
    \foreach \x in {0,...,20} {
 
        \draw ({0.5*\x},0) circle (0.04);
 
    }
 
    \foreach \a in {-3,...,3} {
 
        \draw (10,1)+({\a*20}:1) circle (0.04);
 
        \draw (0,1)+({180+\a*20}:1) circle (0.04);
 
    }
 
    % Numbers
 
    \foreach \x in {0,...,20} {
 
        \draw ({0.5*\x},0.3) node {$\x$};
 
    }
 
    % Patch P
 
    \draw[red] (0.5*6,-0.5) -- (0.5*14,-0.5);
 
    \foreach \x in {6,...,14} {
 
        \draw[fill,red]  ({0.5*\x},-0.5) circle (0.05);
 
    }
 
    % S cap P
 
    \foreach \x in {7,8,10,11,13} {
 
        \draw[fill,blue] ({0.5*\x},-1.0)+(-0.05,-0.05) rectangle +(0.05,0.05);
 
    }
 
    % S \ P
 
    \foreach \x in {1,2,4,16,17,19} {
 
        \draw ({0.5*\x},-1.0) circle (0.07);
 
    }
 
    \draw (3,1) node {$P_\mathrm{min}$};
 
    \draw (7,1) node {$P_\mathrm{max}$};
 
    \draw (0.5*1,1) node {$S_\mathrm{min}$};
 
    \draw (0.5*19,1) node {$S_\mathrm{max}$};
 
    \draw[->] (3,0.8) -- +(0,-0.3);
 
    \draw[->] (7,0.8) -- +(0,-0.3);
 
    \draw[->] (0.5*1,0.8) -- +(0,-0.3);
 
    \draw[->] (0.5*19,0.8) -- +(0,-0.3);
 

	
 
    \draw (0.5*2.5,-1.5) node {$\leftarrow S_\mathrm{left} \rightarrow$};
 
    \draw (0.5*17.5,-1.5) node {$\leftarrow S_\mathrm{right} \rightarrow$};
 

	
 
    % Rectangle around legend
 
    \draw (10.7,-0.2) rectangle (12.3,-1.8);
 

	
 
    \draw[fill,red] (11,-0.5) circle (0.05);
 
    \draw (11.1,-0.5) node[anchor=west] {$P$};
 

	
 
    \draw[fill,blue] (11,-1.0)+(-0.05,-0.05) rectangle +(0.05,0.05);
 
    \draw (11.1,-1.0) node[anchor=west] {$S\cap P$};
 

	
 
    \draw (11,-1.5) circle (0.07);
 
    \draw (11.1,-1.5) node[anchor=west] {$S\cap\overline{P}$};
 
\end{tikzpicture}
 

	
 
\end{document}
main.tex
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@@ -558,110 +558,157 @@ The intuition of the following lemma is that the far right can only affect the z
 
        P_{I'_{>k}}(\mathrm{NZ}^{(k)}) +\mathcal{O}(p^{I_{\max}+1-|I|})	
 
		\tag{as $P_{I'_{<k}}(Z^{(0)}_k)=\mathcal{O}(p^{k-|I'_{<k}|})$}\\
 
		&=\sum_{k\in\mathbb{N}\setminus I}P_{I'_{<k}}(Z^{(0)}_k)\cdot
 
        P_{I'_{>k}}(\mathrm{NZ}^{(k)}) +\mathcal{O}(p^{I_{\max}+1-|I|})\\
 
		&=\sum_{k\in\mathbb{N}\setminus I'}P_{I'_{<k}}(Z^{(0)}_k)\cdot
 
        P_{I'_{>k}}(\mathrm{NZ}^{(k)}) +\mathcal{O}(p^{I_{\max}+1-|I|})	\tag{$k=I_{\max}\Rightarrow P_{I'_{<k}}(Z^{(0)}_k)=\mathcal{O}(p^{I_{\max}-|I'|})=\mathcal{O}(p^{I_{\max}+1-|I|})$}\\
 
		&=P_{I'}(Z^{(0)}) +\mathcal{O}(p^{I_{\max}+1-|I|})	\tag{analogously to the beginning}			
 
	\end{align*}
 
\end{proof}
 

	
 
	The main insight that Lemma~\ref{lemma:probIndep} gives is that if we separate the slots to two halves, in order to see the cancellation of the contribution of the expected resamples on the right, we can simply pair up the left configurations by the particle filling the leftmost slot. And similarly for cancelling the left expectations we pair up right configurations based on the rightmost filling. 
 
	
 
	Also this claim finally ``sees'' how many empty places are between slots. These properties make it possible to use this lemma to prove the sought linear bound. We show it for the infinite chain, but with a little care it should also translate to the circle.
 

	
 
\begin{definition}[Connected patches]
 
	Let $\mathcal{P}\subset 2^{\mathbb{Z}}$ be a finite system of finite subsets of $\mathbb{Z}$. We say that the patch set of a resample sequence is $\mathcal{P}$,
 
	if the connected components of the vertices that have ever become $0$ are exactly the elements of $\mathcal{P}$. We denote by $A^{(\mathcal{P})}$ the event that the set of patches is $\mathcal{P}$. For a patch $P$ let $A^{(P)}=\bigcup_{\mathcal{P}:P\in \mathcal{P}}A^{(\mathcal{P})}$.
 
\end{definition} 
 
Note by Tom: So $A^{(\mathcal{P})}$ is the event that the set of all patches is \emph{exactly} $\mathcal{P}$ whereas $A^{(P)}$ is the event that one of the patches is equal to $P$ but there can be other patches as well.
 

	
 
\begin{definition}[Conditional expectations]
 
	Let $S\subset\mathbb{Z}$ be a finite slot configuration, and for $f\in\{0,1'\}^{|S|}$ let $I:=S(f)$ be the set of vertices filled with particles. 
 
	Then we define
 
	$$R_I:=\mathbb{E}[\#\{\text{resamplings when started from inital state }I\}].$$
 
	For a patch set $\mathcal{P}$ and some $P\in\mathcal{P}$ we define
 
	$$R^{(\mathcal{P})}_I:=\mathbb{E}[\#\{\text{resamplings when started from inital state }I\}|A^{(\mathcal{P})}]$$	
 
	and 
 
	$$R^{(P,\mathcal{P})}_I:=\mathbb{E}[\#\{\text{resamplings inside }P\text{ when started from inital state }I\}|A^{(\mathcal{P})}]$$		
 
	finally
 
	$$R^{(P)}_I:=\mathbb{E}[\#\{\text{resamplings inside }P\text{ when started from inital state }I\}|A^{(P)}].$$	
 
\end{definition} 
 

	
 
    Similarly to Mario's proof I use the observation that 
 
    \begin{align*}
 
    R^{(n)} &= \frac{1}{n}\sum_{b\in\{0,1,1'\}^{n}} \rho_b \; R_{\bar{b}}(p)\\
 
    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} R_{S(f)}\\
 
    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)}
 
    \sum_{\mathcal{P}\text{ patches}} \mathbb{P}_{S(f)}(A^{(\mathcal{P})}) R^{(\mathcal{P})}_{S(f)} \\
 
    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)}
 
    \sum_{\mathcal{P}\text{ patches}} \mathbb{P}_{S(f)}(A^{\mathcal{P}}) \sum_{P\in\mathcal{P}} R^{(P,\mathcal{P})}_{S(f)}\\
 
    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} 
 
    \sum_{\mathcal{P}\text{ patches}} \mathbb{P}_{S(f)}(A^{\mathcal{P}}) \sum_{P\in\mathcal{P}} R^{(P)}_{S(f)\cap P}\tag{by Claim~\ref{claim:eventindependence}}\\ 
 
    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} 
 
    \sum_{P\text{ patch}} R^{(P)}_{S(f)\cap P}\sum_{\mathcal{P}:P\in\mathcal{P}}\mathbb{P}_{S(f)}(A^{\mathcal{P}})\\     
 
    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{P\text{ patch}}\sum_{f\in\{0,1'\}^{|S|}}
 
     \rho_{S(f)} R^{(P)}_{S(f)\cap P}\mathbb{P}_{S(f)}(A^{(P)}) \tag{by definition}\\        
 
    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{P\text{ patch}}\sum_{f\in\{0,1'\}^{|S|}}
 
    \rho_{S(f)} R^{(P)}_{S(f)\cap P}\mathbb{P}_{S(f)\cap P}(A^{(P)})\mathbb{P}_{S(f)\cap \overline{P}}(\overline{Z^{(P_{\min}-1)}}\cap\overline{Z^{(P_{\max}+1)}}) \tag{remember Definition~\ref{def:visitingResamplings} and use Claim~\ref{claim:eventindependence}}\\    
 
    &= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{P\text{ patch}}\sum_{f_P\in\{0,1'\}^{|S\cap P|}}
 
    \rho_{S(f_P)}  R^{(P)}_{S(f_P)}\mathbb{P}_{S(f_P)}(A^{(P)})
 
    \sum_{f_{\overline{P}}\in\{0,1'\}^{|S\cap \overline{P}|}}\rho_{S(f_{\overline{P}})}\mathbb{P}_{S(f_{\overline{P}})}(\overline{Z^{(P_{\min}-1)}}\cap\overline{Z^{(P_{\max}+1)}}) \\   
 
    &= \frac{1}{n}\sum_{S\subseteq [n]} \sum_{P\text{ patch}} \sum_{f_P\in\{0,1'\}^{|S\cap P|}}
 
    \rho_{S(f_P)} R^{(P)}_{S(f_P)} \mathbb{P}_{S(f_P)}(A^{(P)})
 
    \sum_{f_{\overline{P}}\in\{0,1'\}^{|S\cap \overline{P}|}} \rho_{S(f_{\overline{P}})} \mathbb{P}_{S(f_{\overline{P}})}(\overline{Z^{(P_{\min}-1)}}\cap\overline{Z^{(P_{\max}+1)}}) \\   
 
	&= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{P\text{ patch}}\sum_{f_P\in\{0,1'\}^{|S\cap P|}}
 
	\rho_{S(f_P)}
 
	\sum_{f_{\overline{P}}\in\{0,1'\}^{|S\cap \overline{P}|}}\rho_{S(f_{\overline{P}})}\mathcal{O}(p^{|S_{><}|}) \\             
 
        \sum_{f_{\overline{P}}\in\{0,1'\}^{|S\cap \overline{P}|}}\rho_{S(f_{\overline{P}})}\mathcal{O}(p^{|S_{><}|}) \tag{see below} \\
 
	&= \frac{1}{n}\sum_{S\subseteq [n]}\mathcal{O}(p^{|S|+|S_{><}|}).
 
    \end{align*}
 
   	
 
   	The penultimate inequality can be seen by case separation.
 
   	If $S_{><}\subseteq P$ then already $\mathbb{P}_{S(f_P)}(A^{(P)})=\mathcal{O}(p^{|S_{><}|})$.
 
   	Otherwise if all elements of $S_{><}\setminus P$ are larger than $P_{\max}$ then we view the last summation as $\sum_{f'_{\overline{P}}\in\{0,1'\}^{|S\cap \overline{P}\setminus\{S_{\max}\}|}}\sum_{f''_{\overline{P}}\in\{0,1'\}^{1}}$ and use Lemma~\ref{lemma:probIndep} to conclude the cancellations pairwise regarding the filling of $S_{\max}$, i.e., the value of $f''_{\overline{P}}$. We proceed similarly when 
 
   	all elements of $S_{><}\setminus P$ are smaller than $P_{\min}$. In the last case we again proceed similarly, but now the cancellations will come from the interplay of $4$ fillings, corresponding to the possible filling of $S_{\min}$ and $S_{\max}$ simultaneously.
 
   	   
 
\begin{figure}
 
	\begin{center}
 
    	\includegraphics{diagram_patches.pdf}
 
    \end{center}
 
    \caption{\label{fig:patches} Illustration of last steps of the proof.}
 
\end{figure}
 
    The penultimate inequality can be seen by case separation as follows: If $S\subseteq P$ then there is no splitting into $S\cap P$ and $S\setminus P$, and we already have $\mathbb{P}_{S(f_P)}(A^{(P)})=\mathcal{O}(p^{|S_{><}|})$ simply because the patch $P$ must be filled with zeroes that were not yet in $S$, so this is at least $|S_{><}|$ resampled zeroes. For the more general case, assume that $S$ is larger than $P$ on both sides of $P$. This is illustrated in Figure \ref{fig:patches}. We will focus on the following sum that was in the above equations:
 
    \begin{align*}
 
        \sum_{f_{\overline{P}}\in\{0,1'\}^{|S \cap \overline{P}|}} \rho_{S(f_{\overline{P}})} \mathbb{P}_{S(f_{\overline{P}})}(\overline{Z^{(P_{\min}-1)}}\cap\overline{Z^{(P_{\max}+1)}})
 
    \end{align*}
 
    By Lemma \ref{lemma:eventindependence} we can split this sum into two parts: the part to the left of $P$ and the part to the right of $P$. Define $S_\mathrm{left}=S\cap[S_\mathrm{min},P_{\mathrm{min}}-1]$ and $S_\mathrm{right}=S\cap[P_{\mathrm{max}}+1,S_\mathrm{max}]$, so that $S\cap\overline{P} = S_\mathrm{left} \cup S_\mathrm{right}$. These are also illustrated in Figure \ref{fig:patches}. Then we have
 
    \begin{align*}
 
        \mathbb{P}_{S(f_{\overline{P}})}(\overline{Z^{(P_{\min}-1)}}\cap\overline{Z^{(P_{\max}+1)}})
 
        &= \mathbb{P}_{S(f_{\mathrm{left}})}(\overline{Z^{(P_{\min}-1)}}) \;\cdot\; \mathbb{P}_{S(f_{\mathrm{right}})}(\overline{Z^{(P_{\max}+1)}})
 
    \end{align*}
 
    and hence we can split the sum. Without loss of generality we now only consider the `right' part of the sum:
 
    \begin{align*}
 
        \sum_{f\in\{0,1'\}^{|S_\mathrm{right}|}} \rho_{S_\mathrm{right}(f)} \mathbb{P}_{S_\mathrm{right}(f)}(\overline{Z^{(P_{\max}+1)}})
 
    \end{align*}
 
    Now further split this sum over the value of $f$ at position $S_\mathrm{max}$:
 
    \begin{align*}
 
        \sum_{f\in\{0,1'\}^{|S_\mathrm{right}\setminus\{S_\mathrm{max}\}|}} \sum_{f'\in\{0,1'\}}
 
        \rho_{S_\mathrm{right}(f\,f')} \mathbb{P}_{S_\mathrm{right}(f\,f')}(\overline{Z^{(P_{\max}+1)}})
 
    \end{align*}
 
    and we use the definition of $\rho$ for the sum over $f'$:
 
    \begin{align*}
 
         \sum_{f\in\{0,1'\}^{|S_\mathrm{right}\setminus\{S_\mathrm{max}\}|}}
 
        \rho_{S_\mathrm{right}(f)} \left(p \mathbb{P}_{S_\mathrm{right}(f\, 0)}(\overline{Z^{(P_{\max}+1)}}) + (-p) \mathbb{P}_{S_\mathrm{right}(f\, 1)}(\overline{Z^{(P_{\max}+1)}}) \right)
 
    \end{align*}
 
    Now we recognize the setup of Lemma~\ref{lemma:probIndep} with $I=S_\mathrm{right}(f\,0)$ and $I'=S_\mathrm{right}(f\,1)$. The lemma yields
 
    \begin{align*}
 
        \mathbb{P}_{S_\mathrm{right}(f\, 0)}(\overline{Z^{(P_{\max}+1)}}) &= \mathbb{P}_{S_\mathrm{right}(f\, 1)}(\overline{Z^{(P_{\max}+1)}}) + \mathcal{O}(p^{S_\mathrm{max}-(P_{\mathrm{max}}+1)+1-|S_\mathrm{right}|}) \\
 
        &= \mathbb{P}_{S_\mathrm{right}(f\, 1)}(\overline{Z^{(P_{\max}+1)}}) + \mathcal{O}(p^{S_\mathrm{max}-P_{\mathrm{max}}-|S_\mathrm{right}|}) .
 
    \end{align*}
 
    Entering this back into the sum gives
 
    \begin{align*}
 
         \sum_{f\in\{0,1'\}^{|S_\mathrm{right}\setminus\{S_\mathrm{max}\}|}}
 
        \rho_{S_\mathrm{right}(f)} \mathcal{O}(p^{S_\mathrm{max}-P_{\mathrm{max}}-|S_\mathrm{right}|+1})
 
         = \sum_{f\in\{0,1'\}^{|S_\mathrm{right}|}}
 
        \rho_{S_\mathrm{right}(f)} \mathcal{O}(p^{S_\mathrm{max}-P_{\mathrm{max}}-|S_\mathrm{right}|})
 
    \end{align*}
 
    One can do the same for the `left' part, which gives a term $\mathcal{O}(p^{P_\mathrm{min}-S_{\mathrm{min}}-|S_\mathrm{left}|})$. The part of $S$ that was within $P$ gives $\mathbb{P}_{S(f_P)}(A^{(P)})=\mathcal{O}(p^{P_\mathrm{max}-P_\mathrm{min}+1-|S\cap P|})$. Combining these three factors yields
 
    \begin{align*}
 
        (\textrm{left part})(P\textrm{ part})(\textrm{right part}) &=
 
\mathcal{O}(p^{P_\mathrm{min}-S_{\mathrm{min}}-|S_\mathrm{left}|}) \cdot \mathcal{O}(p^{P_\mathrm{max}-P_\mathrm{min}+1-|S\cap P|}) \cdot \mathcal{O}(p^{S_\mathrm{max}-P_{\mathrm{max}}-|S_\mathrm{right}|}) \\
 
        &= \mathcal{O}(p^{S_\mathrm{max}-S_\mathrm{min}+1-|S_\mathrm{left}\cup S_\mathrm{right}\cup (S\cap P)|})\\
 
        &= \mathcal{O}(p^{S_\mathrm{max}-S_\mathrm{min}+1-|S|})
 
        = \mathcal{O}(p^{|S_{><}|})
 
    \end{align*}
 
    as required. This finishes the proof.
 

	
 
    ~
 

	
 
	I think the same arguments would translate to the torus and other translationally invariant spaces, so we could go higher dimensional as Mario suggested. Then I think one would need to replace $|S_{><}|$ by the minimal number $k$ such that there is a $C$ set for which $S\cup C$ is connected. I am not entirely sure how to generalise Lemma~\ref{lemma:probIndep} though, which has key importance in the present proof.
 
    
 
    Questions:
 
    \begin{itemize}
 
    	\item Is this proof finally flawless?
 
    	\item In view of this proof, can we better characterise $a_k^{(k+1)}$?
 
    	\item Why did Mario's and Tom's simulation show that for fixed $C$ the contribution coefficients have constant sign? Is it relevant for proving \ref{it:pos}-\ref{it:geq}?
 
    	\item Can we prove the conjectured formula for $a_k^{(3)}$?		
 
    \end{itemize} 
 
    
 
\begin{comment}
 
    \subsection{Sketch of the (false) proof of the linear bound \ref{it:const}}
 
    Let us interpret $[n]$ as the vertices of a length-$n$ cycle, and interpret operations on vertices mod $n$ s.t. $n+1\equiv 1$ and $1-1\equiv n$.
 
    %\begin{definition}[Resample sequences]
 
    %	A sequence of indices $(r_\ell)=(r_1,r_2,\ldots,r_k)\in[n]^k$ is called resample sequence if our procedure performs $k$ consequtive resampling, where the first resampling of the procedure resamples around the mid point $r_1$ the second around $r_2$ and so on. Let $RS(k)$ the denote the set of length $k$ resample sequences, and let $RS=\cup_{k\in\mathbb{N}}RS(k)$.
 
    %\end{definition}
 
    %\begin{definition}[Constrained resample sequence]\label{def:constrainedRes}
 
    %	Let $C\subseteq[n]$ denote a slot configuration, and let $a\in\{\text{res},\neg\text{res}\}^{n-|C|}$, where the elements correspond to labels ``resampled" vs. ``not resampled" respectively. 
 
    %	For $j\in[n-|C|]$ let $i_j$ denote the $j$-th index in $[n]\setminus C$.
 
    %	We define the set $A^{(C,a)}\subseteq RS$ as the set of resample sequences $(r_\ell)$ such that for all $j$ which has $a_j=\text{res}$ we have that $i_j$ appears in $(r_\ell)$ but for $j'$-s which have $a_{j'}=\neg\text{res}$ we have that $i_{j'}$ never appears in $(r_\ell)$. 
 
    %\end{definition}    
 
    \begin{definition}[Conditional expected number of resamples]
 
    	For a slot configuration $C\subseteq[n]$ and $a\in\{\!\text{ever},\text{ never}\}^{n-|C|}$ we define the event $A^{(C,a)}:=\bigwedge_{j\in[n-|C|]}\{i_j\text{ has }a_j\text{ become }0\text{ before reaching }\mathbf{1}\}$,
 
    	where $i_j$ is the $j$-th vertex of $[n]\setminus C$.
 
    	Then we also define
 
    	$$R^{(C,a)}_b:=\mathbb{E}[\#\{\text{resamplings when started from inital state }b\}|A^{(C,a)}].$$
 
    \end{definition}     
 
    
 
    As in Mario's proof I use the observation that 
 
    \begin{align*}
 
    R^{(n)}(p) &= \frac{1}{n}\sum_{b\in\{0,1,1'\}^{n}} \rho_b \; R_{\bar{b}}(p)\\
 
    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)}(p)\\
 
    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{f\in\{0,1'\}^{|C|}}\sum_{a\in\{\!\text{ever},\text{ never}\}^{n-|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)P_{C(f)}(A^{(C,a)})\\
 
    &= \frac{1}{n}\sum_{C\subseteq [n]}\sum_{a\in\{\!\text{ever},\text{ never}\}^{n-|C|}} \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)P_{C(f)}(A^{(C,a)}), 
 
    \end{align*}
 
    where we denote by $C\subseteq[n]$ a slot configuration, whereas $C(f)$ denotes the slots of $C$ filled with the particles described by $f$, while all other location in $[n]\setminus C$ are set to $1$. 
 
    When we write $R_{C(f)}$ we mean $R_{C(\bar{f})}$, i.e., replace $1'$-s with $1$-s. Since the notation is already heavy we dropped the bar from $f$, as it is clear from the context. Finally by $P_{C(f)}(A^{(C,a)})$ we denote the probability that the event $A^{(C,a)}$ holds.
 
    
 
    As in Definition for $j\in[n-|C|]$ let $i_j$ denote the $j$-th index in $[n]\setminus C$.
 
    Suppose that $a$ is such that there are two indices $j_1\neq j_2$ such that 
 
    $a_{j_1}=\text{never}=a_{j_2}$, moreover the sets $\{i_{j_1}+1,\ldots, i_{j_2}-1\}$ and $\{i_{j_2}+1,\ldots, i_{j_1}-1\}$ partition $C$ non-trivially, and we denote by $C_l$,$C_r$ the corresponding partitions. 
 
    I wanted to prove that
 
    \begin{equation}\label{eq:conditionalCancellation}
 
		\sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R^{{(C,a)}}_{C(f)}(p)=0,
 
    \end{equation}    
 
    based on the observation that for all $f\in\{0,1'\}^{|C|}$ we have 
 
    that 
 
    \begin{equation}\label{eq:keyIndependce}
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