refined core

9 files changed, 1117 insertions, 583 deletions

diff --git a/chapters/core/01-implementation.tex b/chapters/core/01-implementation.tex
index 3e8fa20..32154ef 100644
--- a/chapters/core/01-implementation.tex
+++ b/chapters/core/01-implementation.tex
@@ -1,7 +1,7 @@
 \section{Implementation}
 \label{sec:implementation}
 
-This sections contains details on the implementation of the reduction engine (\textbf{\Cref{sec:inpla}}),
+This section contains details on the implementation of the reduction engine (\textbf{\Cref{sec:inpla}}),
 the interaction rules designed (\textbf{\Cref{sec:interaction-rules}}), the algorithm used in the
 ONNX-to-IN translation (\textbf{\Cref{sec:translation}}) and the Python module (\textbf{\Cref{sec:python-module}}).
 
diff --git a/chapters/core/03-benchmarks.tex b/chapters/core/03-benchmarks.tex
index 40c1986..132b56d 100644
--- a/chapters/core/03-benchmarks.tex
+++ b/chapters/core/03-benchmarks.tex
@@ -1,4 +1,86 @@
 \section{Benchmarks}
 \label{sec:benchmarks}
 
-% This section contains the varius benchmarks
+This section contains the benchmark data, illustrated in \textbf{\Cref{tab:benchmarks}}.
+
+VEIN was tested on a laptop Zenbook UX3402ZA with:
+\begin{itemize}
+	\item \textbf{CPU}: 12th Gen Intel Core i5-1240P
+	\item \textbf{RAM}: 8GiB
+	\item \textbf{SWAP}: 16GiB on NVMe
+\end{itemize}
+using Hyperfine\footnote{\href{https://github.com/sharkdp/hyperfine}{\textbf{Hyperfine Github repository}}.}
+benchmarking tool. Hyperfine runs the test ten times, then reports the mean and standard deviation of
+execution time.
+
+\paragraph{Iris}
+Dataset of iris flowers presented by Fisher et al.~\cite{fisher1936iris}. It consists of three
+species of flower (\textit{Iris setosa}, \textit{Iris virginica} and \textit{Iris versicolor}) with fifty samples each. Each
+sample has four features: width and length of sepals and petals.
+\paragraph{Pendulum}
+A neural safety certificate for stabilizing a pendulum to its upright position under a bound on its
+angle $\theta$, with state $x=[\theta,\dot\theta]\in\mathbb{R}^2$ and nonlinear dynamics.
+\paragraph{Double Integrator}
+A neural safety certificate for stabilizing a second-order linear system to the origin under a bound
+on its position $p$, with state $x=[p,\dot p]\in\mathbb{R}^2$.
+\paragraph{MNIST}
+Widely known dataset of handwritten digits presented by LeCun et al.~\cite{lecun2010mnist}. It
+consists of 60'000 training images and 10'000 testing images.
+
+Pendulum and Double Integrator are taken from \textit{cersyve}, a benchmark presented by Kaulen et
+al.~\cite{kaulen20256thinternationalverificationneural}. We check equivalence between a pre-trained
+and a fine-tuned network for each task.
+
+In the Iris and MNIST dataset, we tested equivalence between a NN and its ad-hoc transformation,
+such as pruning always positive or negative ReLUs, as presented by Kumar et al.~\cite{kumar2019equivalentapproximatetransformationsdeep},
+or changing the architecture from wide to deep and viceversa, as presented by Fan et al.~\cite{JMLR:v24:21-0579}.
+
+For two NN $F: \mathbb{R}^n \to \mathbb{R}^m$ and $F': \mathbb{R}^n \to \mathbb{R}^m$, Eleftheriadis et al.~\cite{eleftheriadis2022equivalence}
+define three kinds of equivalence:
+\begin{itemize}
+	\item \textbf{Strict equivalence}: $\forall x \in \mathbb{R}^n \quad F(x) = F'(x)$.
+	\item \textbf{Epsilon equivalence}: $\forall x \in \mathbb{R}^n \quad ||F(x) - F'(x)|| < \epsilon$ for some $\epsilon > 0$.
+	\item \textbf{Argmax equivalence}: $\forall x \in \mathbb{R}^n \quad \mathtt{argmax}(F(x))=\mathtt{argmax}(F'(x))$ where \texttt{argmax}
+		is the function that returns the index of the maximum value of a vector.
+\end{itemize}
+
+\begin{table}[H]
+	\centering
+	\begin{tabular}[t]{ccccc}
+		\hline
+		Benchmark & Equivalence & Direct [s] & VEIN [s] & Status \\
+		\hline
+		Iris (Stably Active) & Strict & 0.0 & 0.0 & X \\
+		Iris (Stably Active) & Epsilon & 0.0 & 0.0 & X \\
+		Iris (Stably Active) & Argmax & 0.0 & 0.0 & X \\
+		Iris (Stably Inactive) & Strict & 0.0 & 0.0 & X \\
+		Iris (Stably Inactive) & Epsilon & 0.0 & 0.0 & X \\
+		Iris (Stably Inactive) & Argmax & 0.0 & 0.0 & X \\
+		Iris (To Wide) & Strict & 0.0 & 0.0 & X \\
+		Iris (To Wide) & Epsilon & 0.0 & 0.0 & X \\
+		Iris (To Wide) & Argmax & 0.0 & 0.0 & X \\
+		Iris (To Deep) & Strict & 0.0 & 0.0 & X \\
+		Iris (To Deep) & Epsilon & 0.0 & 0.0 & X \\
+		Iris (To Deep) & Argmax & 0.0 & 0.0 & X \\
+		Pendulum & Strict & 0.0 & 88.427 ± 0.670 & SAT \\
+		Pendulum & Epsilon & 0.0 & 0.0 & X \\
+		Pendulum & Argmax & 0.0 & 0.0 & X \\
+		Double Integrator & Strict & 0.0 & 0.0 & X \\
+		Double Integrator & Epsilon & 0.0 & 0.0 & X \\
+		Double Integrator & Argmax & 0.0 & 0.0 & X \\
+		MNIST (Stably Active) & Strict & 0.0 & 0.0 & X \\
+		MNIST (Stably Active) & Epsilon & 0.0 & 0.0 & X \\
+		MNIST (Stably Active) & Argmax & 0.0 & 0.0 & X \\
+		MNIST (Stably Inactive) & Strict & 0.0 & 0.0 & X \\
+		MNIST (Stably Inactive) & Epsilon & 0.0 & 0.0 & X \\
+		MNIST (Stably Inactive) & Argmax & 0.0 & 0.0 & X \\
+		MNIST (To Wide) & Strict & 0.0 & 0.0 & X \\
+		MNIST (To Wide) & Epsilon & 0.0 & 0.0 & X \\
+		MNIST (To Wide) & Argmax & 0.0 & 0.0 & X \\
+		MNIST (To Deep) & Strict & 0.0 & 0.0 & X \\
+		MNIST (To Deep) & Epsilon & 0.0 & 0.0 & X \\
+		MNIST (To Deep) & Argmax & 0.0 & 0.0 & X \\
+	\end{tabular}
+	\caption{Benchmark table.}
+	\label{tab:benchmarks}
+\end{table}
diff --git a/chapters/core/implementation/01-inpla.tex b/chapters/core/implementation/01-inpla.tex
index 5b2eab9..eeebf47 100644
--- a/chapters/core/implementation/01-inpla.tex
+++ b/chapters/core/implementation/01-inpla.tex
@@ -1,7 +1,7 @@
-\subsection{Inpla fork}
+\subsection{INPLA fork}
 \label{sec:inpla}
 
-The  VEIN reduction engine utilizes a modified version of INPLA\footnote{\textbf{\href{https://github.com/inpla/inpla/blob/main/Gentle_introduction_Inpla.md}{INPLA documentation.}}},
+The VEIN reduction engine utilizes a modified version of INPLA\footnote{\textbf{\href{https://github.com/inpla/inpla/blob/main/Gentle_introduction_Inpla.md}{INPLA documentation.}}},
 a multi-threaded parallel interpreter of IN. Performance and ease of use motivated the choice of
 INPLA. Additionally, it supports attribute values, which is a special extension that allows agents
 to hold numerical values at their ports.
@@ -18,38 +18,48 @@ Several modifications adapt INPLA to the pipeline of the framework:
 \end{itemize}
 
 \paragraph{INPLA syntax}
-INPLA evaluates nets, they consist of connections between terms. Terms are built on names and agents:
+INPLA evaluates nets which consist of connections between terms. Terms are built on names and agents:
+\begin{small}
 \begin{verbatim}
 <term> ::= <name> | <agent>
 <name> ::= <nameID>
 <agent> ::= <agentID>
-          | <agentID> ['(' <term> ',' ... ',' <term> ')']
+					| <agentID> ['(' <term> ',' ... ',' <term> ')']
 \end{verbatim}
+\end{small}
 \begin{itemize}
 	\item \textbf{Name}: it works as a buffer between terms.
 	\item \textbf{Agent}: it works as a constructor and de-constructor (defined functions).
 \end{itemize}
 A connection is a relation between two terms and the symbol \texttt{\~} expresses this relation.
 Interaction rules rewrite connections between agents:
+\begin{small}
 \begin{verbatim}
 <interaction-rule> ::= <rule-agent> '><' <rule-agent> '=>' <connections> ';'
 <rule-agent> ::= <agentID>
-               | <agentID> '(' <name> ',' ... ',' <name> ')'
+							 | <agentID> '(' <name> ',' ... ',' <name> ')'
 \end{verbatim}
+\end{small}
 
 \paragraph{Example}
 Unary natural numbers are built by $\texttt{Z}$ (zero) and $\texttt{S}$ (successive function). For
 instance, 0, 1, 2, 3 are expressed as \texttt{Z}, \texttt{S(Z)}, \texttt{S(S(Z))}, \texttt{S(S(S(Z)))}. Here, let's think about an
-increment operation "inc" such that:
+increment operation ``inc'' such that:
+\begin{small}
 \begin{verbatim}
 inc(n) = S(n).
 \end{verbatim}
+\end{small}
 This is written as the following rules:
+\begin{small}
 \begin{verbatim}
 inc(r) >< Z => r ~ S(Z);
 inc(r) >< S(x) => r ~ S(S(x));
 \end{verbatim}
+\end{small}
 Then, the result of \texttt{inc(r) \~{} S(S(Z))} is:
+\begin{small}
 \begin{verbatim}
 r ~ S(S(S(Z)));
 \end{verbatim}
+\end{small}
diff --git a/chapters/core/implementation/02-interaction-rules.tex b/chapters/core/implementation/02-interaction-rules.tex
index b657ee9..49a8a70 100644
--- a/chapters/core/implementation/02-interaction-rules.tex
+++ b/chapters/core/implementation/02-interaction-rules.tex
@@ -10,13 +10,17 @@ The agents, illustrated in \textbf{\Cref{fig:agents}}, can be categorized into f
 			\item \textbf{Concrete}: Represents a constant value $k$.
 		\end{itemize}
 	\item
-		\textbf{Operators}: Agents that perform some kind of operation on the carrier agents.
+		\textbf{Computational Operators}: Agents that perform some kind of operation on the carrier agents.
 		\begin{itemize}
 			\item \textbf{Add}: Binary operator for addition.
 			\item \textbf{Mul}: Binary operator for multiplication.
 			\item \textbf{ReLU}: Unary operator for rectified linear unit.
+		\end{itemize}
+	\item
+		\textbf{Structural Operators}: Agents that perform some kind of operation on net structure.
+		\begin{itemize}
 			\item \textbf{Eraser}: Built-in INPLA unary operator to delete other agents.
-			\item \textbf{Duplicator}: Built-in INPLA unary operator to duplicate other agents.
+			\item \textbf{Dup}: Built-in INPLA unary operator to duplicate other agents.
 		\end{itemize}
 	\item
 		\textbf{Intermediates}: Agents needed to perform each step of the operators.
@@ -28,32 +32,37 @@ The agents, illustrated in \textbf{\Cref{fig:agents}}, can be categorized into f
 			\item \textbf{Materialize}: Converts a Linear agent into an explicit representation for the SMT solver.
 		\end{itemize}
 	\item
-		\textbf{Inerts}: Agents that compose the normal form.
+		\textbf{Terminals}: Agents that compose the normal form.
 		\begin{itemize}
 			\item \textbf{TermAdd}: This agent is parsed as addition for the SMT solver.
 			\item \textbf{TermMul}: This agent is parsed as multiplication for the SMT solver.
 			\item \textbf{TermReLU}: This agent is parsed as rectified linear unit for the SMT solver.
-			\item \textbf{Symbolic}: This agent is parsed as specific variable for the SMT solver.
+			\item \textbf{TermSymbolic}: This agent is parsed as a specific variable for the SMT solver.
+			\item \textbf{TermConcrete}: This agent is parsed as a real value for the SMT solver.
 		\end{itemize}
 \end{itemize}
 
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
-		\begin{tabular}{cccc}
+		\begin{tabular}{ccccc}
 			% Linear
+			&
 			\begin{tikzpicture}
 				\agentLinear{L}{q}{r}
 				\inetwirefree(L.pal) \inetwirefree(L.pax)
 				\node[below] at (L.above pal) {$\mathit{out}$};
 				\node[above] at (L.above pax) {$\mathit{x}$};
 			\end{tikzpicture} &
+			&
 			% Concrete
 			\begin{tikzpicture}
 				\agentConcrete{C}{k}
 				\inetwirefree(C.pal)
 				\node[below] at (C.above pal) {$\mathit{out}$};
-			\end{tikzpicture} &
+			\end{tikzpicture} & \\
+			& (a) $\mathit{Linear}$ & & (b) $\mathit{Concrete}$ & \\[0.5cm]
+
 			% Add
 			\begin{tikzpicture}
 				\agentAdd{A}
@@ -69,9 +78,7 @@ The agents, illustrated in \textbf{\Cref{fig:agents}}, can be categorized into f
 				\node[below] at (M.above pal) {$\mathit{a}$};
 				\node[above] at (M.above pax 1) {$\mathit{out}$};
 				\node[above] at (M.above pax 2) {$\mathit{b}$};
-			\end{tikzpicture} \\
-			(a) $\mathit{Linear}$ & (b) $\mathit{Concrete}$ & (c) $\mathit{Add}$ & (d) $\mathit{Mul}$ \\[0.5cm]
-
+			\end{tikzpicture} &
 			% ReLU
 			\begin{tikzpicture}
 				\agentReLU{R}
@@ -79,13 +86,6 @@ The agents, illustrated in \textbf{\Cref{fig:agents}}, can be categorized into f
 				\node[below] at (R.above pal) {$\mathit{x}$};
 				\node[above] at (R.above pax) {$\mathit{out}$};
 			\end{tikzpicture} &
-			% Materialize
-			\begin{tikzpicture}
-				\agentMaterialize{MAT}
-				\inetwirefree(MAT.pal) \inetwirefree(MAT.pax)
-				\node[below] at (MAT.above pal) {$\mathit{x}$};
-				\node[above] at (MAT.above pax) {$\mathit{out}$};
-			\end{tikzpicture} &
 			% Eraser
 			\begin{tikzpicture}
 				\agentEraser{E}
@@ -100,7 +100,7 @@ The agents, illustrated in \textbf{\Cref{fig:agents}}, can be categorized into f
 				\node[above] at (D.above pax 1) {$\mathit{d_{1}}$};
 				\node[above] at (D.above pax 2) {$\mathit{d_{2}}$};
 			\end{tikzpicture} \\
-			(e) $\mathit{ReLU}$ & (f) $\mathit{Materialize}$ & (g) $\mathit{Eraser}$ & (h) $\mathit{Duplicator}$ \\[0.5cm]
+			(c) $\mathit{Add}$ & (d) $\mathit{Mul}$ & (e) $\mathit{ReLU}$ & (f) $\mathit{Eraser}$ & (g) $\mathit{Duplicator}$ \\[0.5cm]
 
 			% AddCheckLinear
 			\begin{tikzpicture}
@@ -131,8 +131,15 @@ The agents, illustrated in \textbf{\Cref{fig:agents}}, can be categorized into f
 				\inetwirefree(MC.pal) \inetwirefree(MC.pax)
 				\node[below] at (MC.above pal) {$\mathit{b}$};
 				\node[above] at (MC.above pax) {$\mathit{out}$};
+			\end{tikzpicture} &
+			% Materialize
+			\begin{tikzpicture}
+				\agentMaterialize{MAT}
+				\inetwirefree(MAT.pal) \inetwirefree(MAT.pax)
+				\node[below] at (MAT.above pal) {$\mathit{x}$};
+				\node[above] at (MAT.above pax) {$\mathit{out}$};
 			\end{tikzpicture} \\
-			(i) $\mathit{AddCheckLinear}$ & (j) $\mathit{MulCheckLinear}$ & (k) $\mathit{AddCheckConcrete}$ & (l) $\mathit{MulCheckConcrete}$ \\[0.5cm]
+			(h) $\mathit{AddCheckLinear}$ & (i) $\mathit{MulCheckLinear}$ & (j) $\mathit{AddCheckConcrete}$ & (k) $\mathit{MulCheckConcrete}$ & (l) $\mathit{Materialize}$ \\[0.5cm]
 
 			% TermAdd
 			\begin{tikzpicture}
@@ -157,13 +164,19 @@ The agents, illustrated in \textbf{\Cref{fig:agents}}, can be categorized into f
 				\node[below] at (TR.above pal) {$\mathit{out}$};
 				\node[above] at (TR.above pax) {$\mathit{x}$};
 			\end{tikzpicture} &
-			% Symbolic
+			% TermSymbolic
 			\begin{tikzpicture}
-				\agentSymbolic{S}{id}
+				\agentTermSymbolic{S}{id}
 				\inetwirefree(S.pal)
 				\node[below] at (S.above pal) {$\mathit{out}$};
+			\end{tikzpicture} &
+			% TermConcrete
+			\begin{tikzpicture}
+				\agentTermConcrete{C}{k}
+				\inetwirefree(C.pal)
+				\node[below] at (C.above pal) {$\mathit{out}$};
 			\end{tikzpicture} \\
-			(m) $\mathit{TermAdd}$ & (n) $\mathit{TermMul}$ & (o) $\mathit{TermReLU}$ & (p) $\mathit{Symbolic}$ \\[0.5cm]
+			(m) $\mathit{TermAdd}$ & (n) $\mathit{TermMul}$ & (o) $\mathit{TermReLU}$ & (p) $\mathit{TermSymbolic}$ & (q) $\mathit{TermConcrete}$ \\[0.5cm]
 		\end{tabular}
 	}
 	\caption{Agents used in VEIN.}
@@ -178,7 +191,7 @@ operators need to check one operand a time. For this reason \textit{Add} and \te
 principal port faces the other operand.
 
 % Linear >< Add
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\interactionrule{
 		\agentLinear{L}{q}{r}[90]
@@ -200,7 +213,7 @@ principal port faces the other operand.
 \end{figure}
 
 % Linear >< Mul
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\interactionrule{
 		\agentLinear{L}{q}{r}[90]
@@ -230,7 +243,7 @@ This rewiring is also implemented when multiplying a \textit{Concrete} with $k=1
 before it is even invoked.
 
 % Concrete >< Add
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\begin{tabular}{c}
@@ -270,7 +283,7 @@ before it is even invoked.
 \end{figure}
 
 % Concrete >< Mul
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\begin{tabular}{c}
@@ -338,7 +351,7 @@ to a TermAdd or TermMul, for Add and Mul respectively, and then to a \textit{Lin
 $q=1,r=0$.
 
 % Linear >< AddCheckLinear
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\begin{tabular}{c}
@@ -426,7 +439,7 @@ $q=1,r=0$.
 \end{figure}
 
 % Linear >< MulCheckLinear
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\begin{tabular}{c}
@@ -485,7 +498,7 @@ For multiplication (\textbf{\Cref{fig:rule-concrete-mulchecklinear}}) the system
 the \textit{Concrete} agent to both the attributes of the \textit{Linear}
 
 % Concrete >< AddCheckLinear
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\interactionrule{
 		\agentConcrete{C}{j}[90]
@@ -505,7 +518,7 @@ the \textit{Concrete} agent to both the attributes of the \textit{Linear}
 \end{figure}
 
 % Concrete >< MulCheckLinear
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\interactionrule{
 		\agentConcrete{C}{j}[90]
@@ -529,7 +542,7 @@ In the rules shown in \textbf{\Cref{fig:rule-linear-addcheckconcrete}} and \text
 we follow the exact same logic except the first operand is a \textit{Concrete} and the second is a \textit{Linear}.
 
 % Linear >< AddCheckConcrete
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\interactionrule{
 		\agentLinear{L}{s}{t}[90]
@@ -549,7 +562,7 @@ we follow the exact same logic except the first operand is a \textit{Concrete} a
 \end{figure}
 
 % Linear >< MulCheckConcrete
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\interactionrule{
 		\agentLinear{L}{s}{t}[90]
@@ -574,7 +587,7 @@ or short-circuited as illustrated in \textbf{\Cref{fig:rule-concrete-addcheckcon
 \textbf{\Cref{fig:rule-concrete-mulcheckconcrete}} following a similar logic to the previous rules.
 
 % Concrete >< AddCheckConcrete
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\begin{tabular}{c}
@@ -610,7 +623,7 @@ or short-circuited as illustrated in \textbf{\Cref{fig:rule-concrete-addcheckcon
 \end{figure}
 
 % Concrete >< MulCheckConcrete
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\begin{tabular}{c}
@@ -659,7 +672,7 @@ or short-circuited as illustrated in \textbf{\Cref{fig:rule-concrete-addcheckcon
 \end{figure}
 
 \paragraph{Linear/Concrete with ReLU}
-When the \textit{Linear} carrier agent interacts with the \textit{ReLU} operator agents there is no
+When the \textit{Linear} carrier agent interacts with the \textit{ReLU} operator agents there is not
 enough information to perform any simplification so, as shown in \textbf{\Cref{fig:rule-linear-relu}},
 the agent is just materialized, passed to a \textit{TermReLU} and then wrapped in a \textit{Linear}.
 In \textbf{\Cref{fig:rule-concrete-relu}} is illustrated that when the \textit{Concrete} carrier agent
@@ -667,7 +680,7 @@ meets the \textit{ReLU} agent, a \textit{Concrete} is wired to the output either
 if $k>0$ or with attribute equal $0$ otherwise.
 
 % Linear >< ReLU
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\interactionrule{
@@ -693,7 +706,7 @@ if $k>0$ or with attribute equal $0$ otherwise.
 \end{figure}
 
 % Concrete >< ReLU
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\begin{tabular}{c}
@@ -729,13 +742,13 @@ if $k>0$ or with attribute equal $0$ otherwise.
 \end{figure}
 
 \paragraph{Linear/Concrete with Materialize}
-When a \textit{Linear} is materialized it explicitly build an AST using \textit{TermAdd} and \textit{TermMul}
+When a \textit{Linear} is materialized it explicitly build an AST using \textit{TermAdd}, \textit{TermMul} and \textit{TermConcrete}
 to recreate $q*x+r$ as shown in \textbf{\Cref{fig:rule-linear-materialize}}.
-When a \textit{Concrete} needs to be materialized nothing needs to be done and it is directly wired to output
+When a \textit{Concrete} needs to be materialized it is converted into \textit{TermConcrete} as illustrated in
 \textbf{\Cref{fig:rule-concrete-materialize}}.
 
 % Linear >< Materialize
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\begin{tabular}{c}
@@ -750,8 +763,8 @@ When a \textit{Concrete} needs to be materialized nothing needs to be done and i
 			}{
 				\agentTermAdd{TA}[90]
 				\agentTermMul[left=of TA.pax 1]{TM}[90]
-				\agentConcrete[left=of TA.pax 2, below=of TM]{C1}{r}[90]
-				\agentConcrete[left=of TM.pax 1]{C2}{q}[90]
+				\agentTermConcrete[left=of TA.pax 2, below=of TM]{C1}{r}[90]
+				\agentTermConcrete[left=of TM.pax 1]{C2}{q}[90]
 				\inetwire(TA.pax 1)(TM.pal) \inetwire(TA.pax 2)(C1.pal) \inetwire(TM.pax 1)(C2.pal)
 				\inetwirefree(TA.pal) \inetwirefree(TM.pax 2)
 				\node [left] at (TM.above pax 2) {$\mathit{x}$};
@@ -767,7 +780,7 @@ When a \textit{Concrete} needs to be materialized nothing needs to be done and i
 				\node [left] at (L.above pax) {$\mathit{x}$};
 				\node [right] at (MAT.above pax) {$\mathit{out}$};
 			}{
-				\agentConcrete{C}{r}[90]
+				\agentTermConcrete{C}{r}[90]
 				\agentEraser[below=of C]{E}[90]
 				\inetwirefree(C.pal) \inetwirefree(E.pal)
 				\node [right] at (C.above pal) {$\mathit{out}$};
@@ -798,7 +811,7 @@ When a \textit{Concrete} needs to be materialized nothing needs to be done and i
 				\node [right] at (MAT.above pax) {$\mathit{out}$};
 			}{
 				\agentTermAdd{TA}[90]
-				\agentConcrete[left=of TA.right pax]{C}{r}[90]
+				\agentTermConcrete[left=of TA.right pax]{C}{r}[90]
 				\inetwire(C.pal)(TA.pax 2)
 				\inetwirefree(TA.pax 1) \inetwirefree(TA.pal)
 				\node [left] at (TA.above pax 1) {$\mathit{x}$};
@@ -815,7 +828,7 @@ When a \textit{Concrete} needs to be materialized nothing needs to be done and i
 				\node [right] at (MAT.above pax) {$\mathit{out}$};
 			}{
 				\agentTermMul{TM}[90]
-				\agentConcrete[left=of TM.left pax]{C}{q}[90]
+				\agentTermConcrete[left=of TM.left pax]{C}{q}[90]
 				\inetwire(C.pal)(TM.pax 1)
 				\inetwirefree(TM.pax 2) \inetwirefree(TM.pal)
 				\node [left] at (TM.above pax 2) {$\mathit{x}$};
@@ -828,7 +841,7 @@ When a \textit{Concrete} needs to be materialized nothing needs to be done and i
 \end{figure}
 
 % Concrete >< Materialize
-\begin{figure}[ht]
+\begin{figure}[H]
 	\centering
 	\resizebox{\textwidth}{!}{
 		\interactionrule{
@@ -838,7 +851,7 @@ When a \textit{Concrete} needs to be materialized nothing needs to be done and i
 			\inetwirefree(MAT.pax)
 			\node [right] at (MAT.above pax) {$\mathit{out}$};
 		}{
-			\agentConcrete{C}{k}[90]
+			\agentTermConcrete{C}{k}[90]
 			\inetwirefree(C.pal)
 			\node [right] at (C.above pal) {$\mathit{out}$};
 		}
diff --git a/chapters/core/implementation/03-translation.tex b/chapters/core/implementation/03-translation.tex
index 638ff17..0a2eec0 100644
--- a/chapters/core/implementation/03-translation.tex
+++ b/chapters/core/implementation/03-translation.tex
@@ -11,15 +11,7 @@ to be able to instantiate the correct number of \textit{Dup} agents. The main al
 in \textbf{\Cref{alg:onnx-to-in}}, maintains an \textit{interactions} dictionary data-structure, that maps each tensor
 name to a list of ports, to keep track of the graph traversal.
 
-To maximize the concurrency of the INPLA engine, the translation layer avoids generating linear
-chains of agents, opting instead for balanced binary trees for signal distribution (single input to
-multiple output) and signal reduction (multiple input to single output). As \textbf{\Cref{alg:balanced-fan-in}}
-and \textbf{\Cref{alg:balanced-fan-out}} illustrate, the depth of agent chains (especially \textit{Dup} chains)
-is limited to $O(\log N)$. The two algorithms are very similar, the difference is in how they wire the
-agents together: in the \textit{Fan-In} the principal port of the agents are facing the leaves, while in the
-\textit{Fan-Out} they are facing the root.
-
-\begin{algorithm}[ht]
+\begin{algorithm}[H]
 	\caption{Backwards ONNX-to-IN Translation}
 	\label{alg:onnx-to-in}
 	\SetKwInOut{Input}{Input}\SetKwInOut{Output}{Output}
@@ -67,22 +59,31 @@ agents together: in the \textit{Fan-In} the principal port of the agents are fac
 					$S.\text{append}(root \sim sink)$
 				}
 			}
-			\Case{Identity}{
-				$interactions[N.input] \leftarrow interactions[N.output]$
-			}
 		}
 	}
 
 	\ForEach{neuron $x$ in $G.input$}{
 		$sink \leftarrow \text{BalancedFanOut}(interactions[G.input][x], \text{Dup}, S)$
 
-		$S.\text{append}(sink \sim \text{Linear}(\text{Symbolic}(x), 1.0, 0.0))$
+		$S.\text{append}(sink \sim \text{Linear}(\text{TermSymbolic}(x), 1.0, 0.0))$
+	}
+
+	\ForEach{$y$ in $\text{len}(interactions[G.output])$}{
+		$S.\text{append}(result_y)$
 	}
 
 	\Return{S}
 \end{algorithm}
 
-\begin{algorithm}[ht]
+To maximize the concurrency of the INPLA engine, the translation layer avoids generating linear
+chains of agents, opting instead for balanced binary trees for signal distribution (single input to
+multiple output) and signal reduction (multiple input to single output). As \textbf{\Cref{alg:balanced-fan-in}}
+and \textbf{\Cref{alg:balanced-fan-out}} illustrate, the depth of agent chains (especially \textit{Dup} chains)
+is limited to $O(\log N)$. The two algorithms are very similar, the difference is in how they wire the
+agents together: in the \textit{Fan-In} the principal port of the agents are facing the leaves, while in the
+\textit{Fan-Out} they are facing the root.
+
+\begin{algorithm}[H]
 	\caption{Balanced Fan-In}
 	\label{alg:balanced-fan-in}
 	\SetKwInOut{Input}{Input}\SetKwInOut{Output}{Output}
@@ -111,8 +112,7 @@ agents together: in the \textit{Fan-In} the principal port of the agents are fac
 	\Return{$T[0]$}
 \end{algorithm}
 
-
-\begin{algorithm}[ht]
+\begin{algorithm}[H]
 	\caption{Balanced Fan-Out}
 	\label{alg:balanced-fan-out}
 	\SetKwInOut{Input}{Input}\SetKwInOut{Output}{Output}
diff --git a/chapters/core/soundness-proof/01-mathematical-definitions.tex b/chapters/core/soundness-proof/01-mathematical-definitions.tex
index 19059ad..4bcb9df 100644
--- a/chapters/core/soundness-proof/01-mathematical-definitions.tex
+++ b/chapters/core/soundness-proof/01-mathematical-definitions.tex
@@ -27,10 +27,18 @@ The agents are defined as:
 		where $x, \mathit{out}$ are wires.
 	\item $\llbracket \mathit{Materialize}(\mathit{out}) \sim x \rrbracket \iff \mathit{out} = x$\\
 		where $x, \mathit{out}$ are wires.
-	\item $\llbracket \mathit{TermAdd}(a, b) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = a + b$\\
+	\item $\llbracket \mathit{TermAdd}(a, b) \sim \mathit{out} \rrbracket \iff \mathit{out} = a + b$\\
 		where $a, b, \mathit{out}$ are wires.
-	\item $\llbracket \mathit{TermMul}(a, b) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = a \cdot b$\\
+	\item $\llbracket \mathit{TermMul}(a, b) \sim \mathit{out} \rrbracket \iff \mathit{out} = a \cdot b$\\
 		where $a, b, \mathit{out}$ are wires.
 	\item $\llbracket \mathit{TermReLU}(x) \sim \mathit{out} \rrbracket \iff \mathit{out} = \max(0, x)$\\
 		where $x, \mathit{out}$ are wires.
+	\item $\llbracket \mathit{TermSymbolic}(id) \sim \mathit{out} \rrbracket \iff \mathit{out} = x_{id}$\\
+		where $\mathit{out}$ is a wire and $id$ is a variable identifier.
+	\item $\llbracket \mathit{TermConcrete}(k) \sim \mathit{out} \rrbracket \iff \mathit{out} = k$\\
+		where $\mathit{out}$ is a wire and $k \in \mathbb{R}$ is an attribute.
+	\item $\llbracket \mathit{Dup}(x, y) \sim z \rrbracket \iff z = x = y$\\
+		where $x, y, z$ are wires.
+	\item $\llbracket \mathit{Eraser} \sim x \rrbracket \iff x \in \mathbb{R}$\\
+		where $x$ is a wire. \textit{Eraser} effectively removes any constraints.
 \end{itemize}
diff --git a/chapters/core/soundness-proof/02-soundness-of-translation.tex b/chapters/core/soundness-proof/02-soundness-of-translation.tex
index 6764ea2..800e085 100644
--- a/chapters/core/soundness-proof/02-soundness-of-translation.tex
+++ b/chapters/core/soundness-proof/02-soundness-of-translation.tex
@@ -3,7 +3,7 @@
 
 We need to prove that for each ONNX operator a semantically equivalent IN is produced.
 
-\paragraph{ReLU}
+\begin{lemma}
 The ONNX ReLU operator for an input tensor X and output tensor Y is defined as:
 $$
 Y = \max(0, X)
@@ -17,8 +17,9 @@ $$
 \llbracket \mathit{ReLU}(y_i) \sim x_i \rrbracket \Rightarrow y_i = \max(0, x_i)
 $$
 Which is identical to the ONNX definition.
+\end{lemma}
 
-\paragraph{Gemm}
+\begin{lemma}
 The ONNX Gemm (General Matrix Multiplication) operator for input tensors A, B, C, input $\alpha$
 and $\beta$ and output tensor Y is defined as:
 $$
@@ -40,15 +41,4 @@ $$
 \end{aligned}
 $$
 By substituting $v_i$, the result matches the ONNX definition.
-
-\paragraph{Identity}
-The ONNX Identity does not modify the numerical values of the tensors. As this operator results in a
-direct wire connection:
-$$
-y_i \sim x_i
-$$
-The semantic:
-$$
-\llbracket y_i \sim x_i \rrbracket \Rightarrow y_i = x_i
-$$
-trivially preserve the identity mapping.
+\end{lemma}
diff --git a/chapters/core/soundness-proof/03-soundness-of-interaction-rules.tex b/chapters/core/soundness-proof/03-soundness-of-interaction-rules.tex
index 2336135..be0870d 100644
--- a/chapters/core/soundness-proof/03-soundness-of-interaction-rules.tex
+++ b/chapters/core/soundness-proof/03-soundness-of-interaction-rules.tex
@@ -1,507 +1,849 @@
 \subsection{Soundness of Interaction Rules}
 \label{sec:soundness-of-interaction-rules}
 
-\paragraph{Linear and Add}
-For the \textit{Linear} and \textit{Add} agents we have the interaction rule:
-$$
-\mathit{Linear}(x, q, r) \bowtie \mathit{Add}(\mathit{out}, b) \Rightarrow \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim b \\
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
-		& \llbracket \mathit{Add}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w + b \\
-		& \Rightarrow \mathit{out} = (q \cdot x + r) + b
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: } & \llbracket \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim b \rrbracket \\
-		& \Rightarrow \mathit{out} = q \cdot x + (r + b)
-	\end{aligned}
-\end{aligned}
-$$
-Since $(q \cdot x + r) + b = q \cdot x + (r + b)$, the rule is sound.
-
-\paragraph{Linear and Mul}
-For the \textit{Linear} and \textit{Mul} agents we have the interaction rule:
-$$
-\mathit{Linear}(x, q, r) \bowtie \mathit{Mul}(\mathit{out}, b) \Rightarrow \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim b \\
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
-		& \llbracket \mathit{Mul}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w \cdot b \\
-		& \Rightarrow \mathit{out} = (q \cdot x + r) \cdot b
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: } & \llbracket \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim b \rrbracket \\
-		& \Rightarrow \mathit{out} = q \cdot b \cdot x + r \cdot b
-	\end{aligned}
-\end{aligned}
-$$
-Since $(q \cdot x + r) \cdot b = q \cdot b \cdot x + r \cdot b$, the rule is sound.
+\begin{lemma}
+	For the \textit{Linear} and \textit{Add} agents we have the interaction rule:
+	$$
+	\mathit{Linear}(x, q, r) \bowtie \mathit{Add}(\mathit{out}, b) \Rightarrow \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim b \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
+			& \llbracket \mathit{Add}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w + b \\
+			& \Rightarrow \mathit{out} = (q \cdot x + r) + b
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim b \rrbracket \\
+			& \Rightarrow \mathit{out} = q \cdot x + (r + b)
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $(q \cdot x + r) + b = q \cdot x + (r + b)$, the rule is sound.
+\end{lemma}
 
-\paragraph{Concrete and Add}
-For the \textit{Concrete} and \textit{Add} agents we have the interaction rule:
-$$
-\mathit{Concrete}(k) \bowtie \mathit{Add}(\mathit{out}, b) \Rightarrow
-\begin{cases}
-	\mathit{out} \sim b & \text{if } k = 0 \\
-	\mathit{AddCheckConcrete}(\mathit{out}, k) \sim b & \text{otherwise}
-\end{cases}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
-		& \llbracket \mathit{Add}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w + b \\
-		& \Rightarrow \mathit{out} = k + b
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: }
-		\begin{cases}
-			\llbracket \mathit{out} \sim b \rrbracket \Rightarrow \mathit{out} = b & \text{if } k = 0 \\
-			\llbracket \mathit{AddCheckConcrete}(\mathit{out}, k) \sim b \rrbracket \Rightarrow \mathit{out} = k + b & \text{otherwise}
-		\end{cases}
+\begin{lemma}
+	For the \textit{Linear} and \textit{Mul} agents we have the interaction rule:
+	$$
+	\mathit{Linear}(x, q, r) \bowtie \mathit{Mul}(\mathit{out}, b) \Rightarrow \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim b \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
+			& \llbracket \mathit{Mul}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w \cdot b \\
+			& \Rightarrow \mathit{out} = (q \cdot x + r) \cdot b
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim b \rrbracket \\
+			& \Rightarrow \mathit{out} = q \cdot b \cdot x + r \cdot b
+		\end{aligned}
 	\end{aligned}
-\end{aligned}
-$$
-Since:
-$$
-\begin{cases}
-	0 + b = b & \text{if } k = 0 \\
-	k + b = k + b & \text{otherwise}
-\end{cases}
-$$
-the rule is sound.
+	$$
+	Since $(q \cdot x + r) \cdot b = q \cdot b \cdot x + r \cdot b$, the rule is sound.
+\end{lemma}
 
-\paragraph{Concrete and Mul}
-For the \textit{Concrete} and \textit{Mul} agents we have the interaction rule:
-$$
-\mathit{Concrete}(k) \bowtie \mathit{Mul}(\mathit{out}, b) \Rightarrow
-\begin{cases}
-	\mathit{out} \sim \mathit{Concrete}(0); b \sim \mathit{Eraser} & \text{if } k = 0 \\
-	\mathit{out} \sim  b & \text{if } k = 1 \\
-	\mathit{MulCheckConcrete}(\mathit{out}, k) \sim b & \text{otherwise}
-\end{cases}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
-		& \llbracket \mathit{Mul}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w \cdot b \\
-		& \Rightarrow \mathit{out} = k \cdot b
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: }
-		\begin{cases}
-			\llbracket \mathit{out} \sim \mathit{Concrete}(0); b \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } k = 0 \\
-			\llbracket \mathit{out} \sim b \rrbracket \Rightarrow \mathit{out} = b & \text{if } k = 1 \\
-			\llbracket \mathit{MulCheckConcrete}(\mathit{out}, k) \sim b \rrbracket \Rightarrow \mathit{out} = k \cdot b & \text{otherwise}
-		\end{cases}
+\begin{lemma}
+	For the \textit{Concrete} and \textit{Add} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(k) \bowtie \mathit{Add}(\mathit{out}, b) \Rightarrow
+	\begin{cases}
+		\mathit{out} \sim b & \text{if } k = 0 \\
+		\mathit{AddCheckConcrete}(\mathit{out}, k) \sim b & \text{otherwise}
+	\end{cases}
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
+			& \llbracket \mathit{Add}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w + b \\
+			& \Rightarrow \mathit{out} = k + b
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+			\begin{cases}
+				\llbracket \mathit{out} \sim b \rrbracket \Rightarrow \mathit{out} = b & \text{if } k = 0 \\
+				\llbracket \mathit{AddCheckConcrete}(\mathit{out}, k) \sim b \rrbracket \Rightarrow \mathit{out} = k + b & \text{otherwise}
+			\end{cases}
+		\end{aligned}
 	\end{aligned}
-\end{aligned}
-$$
-Since:
-$$
-\begin{cases}
-	0 \cdot b = 0 & \text{if } k = 0 \\
-	1 \cdot b = b & \text{if } k = 1 \\
-	k \cdot b = k \cdot b & \text{otherwise}
-\end{cases}
-$$
-the rule is sound.
+	$$
+	Since:
+	$$
+	\begin{cases}
+		0 + b = b & \text{if } k = 0 \\
+		k + b = k + b & \text{otherwise}
+	\end{cases}
+	$$
+	the rule is sound.
+\end{lemma}
 
-\paragraph{Linear and AddCheckLinear}
-For the \textit{Linear} and \textit{AddCheckLinear} agents we have the interaction rule:
-$$
-\begin{aligned}
-	& \mathit{Linear}(y, s, t) \bowtie \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \\
-	& \quad \begin{cases}
-		\mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} & \text{if } q,r,s,t = 0 \\
-		\mathit{out} \sim  \mathit{Linear}(x, q, r); y \sim \mathit{Eraser} & \text{if } s,t = 0 \\
-		\mathit{out} \sim  \mathit{Linear}(y, s, t); x \sim \mathit{Eraser} & \text{if } q,r = 0 \\
-		\begin{aligned}
-			& \mathit{Linear}(x, q, r) \sim \mathit{Materialize}(\mathit{out}_x); \\
-			& \mathit{Linear}(y, s, t) \sim \mathit{Materialize}(\mathit{out}_y); \\
-			& \mathit{out} \sim \mathit{Linear}(\mathit{TermAdd}(\mathit{out}_x, \mathit{out}_y), 1, 0)
+\begin{lemma}
+	For the \textit{Concrete} and \textit{Mul} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(k) \bowtie \mathit{Mul}(\mathit{out}, b) \Rightarrow
+	\begin{cases}
+		\mathit{out} \sim \mathit{Concrete}(0); b \sim \mathit{Eraser} & \text{if } k = 0 \\
+		\mathit{out} \sim  b & \text{if } k = 1 \\
+		\mathit{MulCheckConcrete}(\mathit{out}, k) \sim b & \text{otherwise}
+	\end{cases}
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
+			& \llbracket \mathit{Mul}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w \cdot b \\
+			& \Rightarrow \mathit{out} = k \cdot b
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+			\begin{cases}
+				\llbracket \mathit{out} \sim \mathit{Concrete}(0); b \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } k = 0 \\
+				\llbracket \mathit{out} \sim b \rrbracket \Rightarrow \mathit{out} = b & \text{if } k = 1 \\
+				\llbracket \mathit{MulCheckConcrete}(\mathit{out}, k) \sim b \rrbracket \Rightarrow \mathit{out} = k \cdot b & \text{otherwise}
+			\end{cases}
 		\end{aligned}
-		& \text{otherwise}
+	\end{aligned}
+	$$
+	Since:
+	$$
+	\begin{cases}
+		0 \cdot b = 0 & \text{if } k = 0 \\
+		1 \cdot b = b & \text{if } k = 1 \\
+		k \cdot b = k \cdot b & \text{otherwise}
 	\end{cases}
-\end{aligned}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
-		& \llbracket \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot x + (r + w) \\
-		& \Rightarrow \mathit{out} = q \cdot x + (r + s \cdot y + t)
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: }
-		\begin{cases}
-			\llbracket \mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } q,r,s,t = 0 \\
-			\llbracket \mathit{out} \sim  \mathit{Linear}(x, q, r); y \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = q \cdot x + r & \text{if } s,t = 0 \\
-			\llbracket \mathit{out} \sim  \mathit{Linear}(y, s, t); x \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = s \cdot y + t & \text{if } q,r = 0 \\
+	$$
+	the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{Linear} and \textit{AddCheckLinear} agents we have the interaction rule:
+	$$
+	\begin{aligned}
+		& \mathit{Linear}(y, s, t) \bowtie \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \\
+		& \quad \begin{cases}
+			\mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} & \text{if } q,r,s,t = 0 \\
+			\mathit{out} \sim  \mathit{Linear}(x, q, r); y \sim \mathit{Eraser} & \text{if } s,t = 0 \\
+			\mathit{out} \sim  \mathit{Linear}(y, s, t); x \sim \mathit{Eraser} & \text{if } q,r = 0 \\
 			\begin{aligned}
-				& \llbracket \mathit{Linear}(x, q, r) \sim w_1 \rrbracket \Rightarrow w_1 = q \cdot x + r \\
-				& \llbracket \mathit{Materialize}(\mathit{out}_x) \sim w_1 \rrbracket \Rightarrow \mathit{out}_x = w_1 \\
-				& \llbracket \mathit{Linear}(y, s, t) \sim w_2 \rrbracket \Rightarrow w_2 = s \cdot y + t \\
-				& \llbracket \mathit{Materialize}(\mathit{out}_y) \sim w_2 \rrbracket \Rightarrow \mathit{out}_y = w_2 \\
-				& \llbracket \mathit{Linear}(\mathit{TermAdd}(\mathit{out}_x, \mathit{out}_y), 1, 0) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = 1 \cdot (\mathit{out}_x + \mathit{out}_y) + 0
-			\end{aligned}\\
-			\Rightarrow \mathit{out} = 1 \cdot (q \cdot x + r + s \cdot y + t) + 0 & \text{otherwise}
+				& \mathit{Linear}(x, q, r) \sim \mathit{Materialize}(\mathit{out}_x); \\
+				& \mathit{Linear}(y, s, t) \sim \mathit{Materialize}(\mathit{out}_y); \\
+				& \mathit{out} \sim \mathit{Linear}(\mathit{TermAdd}(\mathit{out}_x, \mathit{out}_y), 1, 0)
+			\end{aligned}
+			& \text{otherwise}
 		\end{cases}
 	\end{aligned}
-\end{aligned}
-$$
-Since:
-$$
-\begin{cases}
-	0 \cdot x + (0 + 0 \cdot y + 0) = 0 & \text{if } q,r,s,t = 0 \\
-	q \cdot x + (r + 0 \cdot y + 0) = q \cdot x + r & \text{if } s,t = 0 \\
-	0 \cdot x + (0 + s \cdot y + t) = s \cdot y + t & \text{if } q,r = 0 \\
-	q \cdot x + (r + s \cdot y + t) = 1 \cdot (q \cdot x + r + s \cdot y + t) + 0 & \text{otherwise}
-\end{cases}
-$$
-the rule is sound.
-
-\paragraph{Linear and MulCheckLinear}
-For the \textit{Linear} and \textit{MulCheckLinear} agents we have the interaction rule:
-$$
-\begin{aligned}
-	& \mathit{Linear}(y, s, t) \bowtie \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \\
-	& \quad \begin{cases}
-		\mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} & \text{if } q,r = 0 \lor s,t = 0 \\
-		\begin{aligned}
-			& \mathit{Linear}(x, q, r) \sim \mathit{Materialize}(\mathit{out}_x); \\
-			& \mathit{Linear}(y, s, t) \sim \mathit{Materialize}(\mathit{out}_y); \\
-			& \mathit{out} \sim \mathit{Linear}(\mathit{TermMul}(\mathit{out}_x, \mathit{out}_y), 1, 0)
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
+			& \llbracket \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot x + (r + w) \\
+			& \Rightarrow \mathit{out} = q \cdot x + (r + s \cdot y + t)
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+			\begin{cases}
+				\llbracket \mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } q,r,s,t = 0 \\
+				\llbracket \mathit{out} \sim  \mathit{Linear}(x, q, r); y \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = q \cdot x + r & \text{if } s,t = 0 \\
+				\llbracket \mathit{out} \sim  \mathit{Linear}(y, s, t); x \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = s \cdot y + t & \text{if } q,r = 0 \\
+				\begin{aligned}
+					& \llbracket \mathit{Linear}(x, q, r) \sim w_1 \rrbracket \Rightarrow w_1 = q \cdot x + r \\
+					& \llbracket \mathit{Materialize}(\mathit{out}_x) \sim w_1 \rrbracket \Rightarrow \mathit{out}_x = w_1 \\
+					& \llbracket \mathit{Linear}(y, s, t) \sim w_2 \rrbracket \Rightarrow w_2 = s \cdot y + t \\
+					& \llbracket \mathit{Materialize}(\mathit{out}_y) \sim w_2 \rrbracket \Rightarrow \mathit{out}_y = w_2 \\
+					& \llbracket \mathit{Linear}(\mathit{TermAdd}(\mathit{out}_x, \mathit{out}_y), 1, 0) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = 1 \cdot (\mathit{out}_x + \mathit{out}_y) + 0
+				\end{aligned}\\
+				\Rightarrow \mathit{out} = 1 \cdot (q \cdot x + r + s \cdot y + t) + 0 & \text{otherwise}
+			\end{cases}
 		\end{aligned}
-		& \text{otherwise}
+	\end{aligned}
+	$$
+	Since:
+	$$
+	\begin{cases}
+		0 \cdot x + (0 + 0 \cdot y + 0) = 0 & \text{if } q,r,s,t = 0 \\
+		q \cdot x + (r + 0 \cdot y + 0) = q \cdot x + r & \text{if } s,t = 0 \\
+		0 \cdot x + (0 + s \cdot y + t) = s \cdot y + t & \text{if } q,r = 0 \\
+		q \cdot x + (r + s \cdot y + t) = 1 \cdot (q \cdot x + r + s \cdot y + t) + 0 & \text{otherwise}
 	\end{cases}
-\end{aligned}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
-		& \llbracket \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot w \cdot x + r \cdot w \\
-		& \Rightarrow \mathit{out} = q \cdot (s \cdot y + t) \cdot x + r \cdot (s \cdot y + t)
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: }
-		\begin{cases}
-			\llbracket \mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } q,r = 0 \lor s,t = 0 \\
+	$$
+	the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{Linear} and \textit{MulCheckLinear} agents we have the interaction rule:
+	$$
+	\begin{aligned}
+		& \mathit{Linear}(y, s, t) \bowtie \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \\
+		& \quad \begin{cases}
+			\mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} & \text{if } q,r = 0 \lor s,t = 0 \\
 			\begin{aligned}
-				& \llbracket \mathit{Linear}(x, q, r) \sim w_1 \rrbracket \Rightarrow w_1 = q \cdot x + r \\
-				& \llbracket \mathit{Materialize}(\mathit{out}_x) \sim w_1 \rrbracket \Rightarrow \mathit{out}_x = w_1 \\
-				& \llbracket \mathit{Linear}(y, s, t) \sim w_2 \rrbracket \Rightarrow w_2 = s \cdot y + t \\
-				& \llbracket \mathit{Materialize}(\mathit{out}_y) \sim w_2 \rrbracket \Rightarrow \mathit{out}_y = w_2 \\
-				& \llbracket \mathit{Linear}(\mathit{TermMul}(\mathit{out}_x, \mathit{out}_y), 1, 0) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = 1 \cdot \mathit{out}_x \cdot \mathit{out}_y + 0
-			\end{aligned}\\
-			\Rightarrow \mathit{out} = 1 \cdot (q \cdot x + r) \cdot (s \cdot y + t) + 0 & \text{otherwise}
+				& \mathit{Linear}(x, q, r) \sim \mathit{Materialize}(\mathit{out}_x); \\
+				& \mathit{Linear}(y, s, t) \sim \mathit{Materialize}(\mathit{out}_y); \\
+				& \mathit{out} \sim \mathit{Linear}(\mathit{TermMul}(\mathit{out}_x, \mathit{out}_y), 1, 0)
+			\end{aligned}
+			& \text{otherwise}
 		\end{cases}
 	\end{aligned}
-\end{aligned}
-$$
-Since:
-$$
-\begin{cases}
-	0 \cdot (s \cdot y + t) \cdot x + 0 \cdot (s \cdot y + t) = 0 \lor q \cdot (0 \cdot y + 0) \cdot x + r \cdot (0 \cdot y + 0) = 0 & \text{if } q,r = 0 \lor s,t = 0  \\
-	1 \cdot (q \cdot x + r) \cdot (s \cdot y + t) + 0 = q \cdot (s \cdot y + t) \cdot x + r \cdot (s \cdot y + t) & \text{otherwise}
-\end{cases}
-$$
-the rule is sound.
-
-\paragraph{Concrete and AddCheckLinear}
-For the \textit{Concrete} and \textit{AddCheckLinear} agents we have the interaction rule:
-$$
-\mathit{Concrete}(j) \bowtie \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \mathit{Linear}(x, q, r + j) \sim \mathit{out} \\
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
-		& \llbracket \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot x + (r + w) \\
-		& \Rightarrow \mathit{out} = q \cdot x + (r + j)
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: } & \llbracket \mathit{Linear}(x, q, r + j) \sim \mathit{out} \rrbracket \\
-		& \Rightarrow \mathit{out} = q \cdot x + (r + j)
-	\end{aligned}
-\end{aligned}
-$$
-Since $q \cdot x + (r + j) = q \cdot x + (r + j)$, the rule is sound.
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
+			& \llbracket \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot w \cdot x + r \cdot w \\
+			& \Rightarrow \mathit{out} = q \cdot (s \cdot y + t) \cdot x + r \cdot (s \cdot y + t)
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+			\begin{cases}
+				\llbracket \mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } q,r = 0 \lor s,t = 0 \\
+				\begin{aligned}
+					& \llbracket \mathit{Linear}(x, q, r) \sim w_1 \rrbracket \Rightarrow w_1 = q \cdot x + r \\
+					& \llbracket \mathit{Materialize}(\mathit{out}_x) \sim w_1 \rrbracket \Rightarrow \mathit{out}_x = w_1 \\
+					& \llbracket \mathit{Linear}(y, s, t) \sim w_2 \rrbracket \Rightarrow w_2 = s \cdot y + t \\
+					& \llbracket \mathit{Materialize}(\mathit{out}_y) \sim w_2 \rrbracket \Rightarrow \mathit{out}_y = w_2 \\
+					& \llbracket \mathit{Linear}(\mathit{TermMul}(\mathit{out}_x, \mathit{out}_y), 1, 0) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = 1 \cdot \mathit{out}_x \cdot \mathit{out}_y + 0
+				\end{aligned}\\
+				\Rightarrow \mathit{out} = 1 \cdot (q \cdot x + r) \cdot (s \cdot y + t) + 0 & \text{otherwise}
+			\end{cases}
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since:
+	$$
+	\begin{cases}
+		0 \cdot (s \cdot y + t) \cdot x + 0 \cdot (s \cdot y + t) = 0 \lor q \cdot (0 \cdot y + 0) \cdot x + r \cdot (0 \cdot y + 0) = 0 & \text{if } q,r = 0 \lor s,t = 0  \\
+		1 \cdot (q \cdot x + r) \cdot (s \cdot y + t) + 0 = q \cdot (s \cdot y + t) \cdot x + r \cdot (s \cdot y + t) & \text{otherwise}
+	\end{cases}
+	$$
+	the rule is sound.
+\end{lemma}
 
-\paragraph{Concrete and MulCheckLinear}
-For the \textit{Concrete} and \textit{MulCheckLinear} agents we have the interaction rule:
-$$
-\mathit{Concrete}(j) \bowtie \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \mathit{Linear}(x, q \cdot j, r \cdot j) \sim \mathit{out} \\
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
-		& \llbracket \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot w \cdot x + r \cdot w \\
-		& \Rightarrow \mathit{out} = q \cdot j \cdot x + r \cdot j
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: } & \llbracket \mathit{Linear}(x, q \cdot j, r \cdot j) \sim \mathit{out} \rrbracket \\
-		& \Rightarrow \mathit{out} = (q \cdot j) \cdot x + (r \cdot j)
-	\end{aligned}
-\end{aligned}
-$$
-Since $q \cdot j \cdot x + r \cdot j = (q \cdot j) \cdot x + (r \cdot j)$, the rule is sound.
+\begin{lemma}
+	For the \textit{Concrete} and \textit{AddCheckLinear} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(j) \bowtie \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \mathit{Linear}(x, q, r + j) \sim \mathit{out} \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
+			& \llbracket \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot x + (r + w) \\
+			& \Rightarrow \mathit{out} = q \cdot x + (r + j)
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Linear}(x, q, r + j) \sim \mathit{out} \rrbracket \\
+			& \Rightarrow \mathit{out} = q \cdot x + (r + j)
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $q \cdot x + (r + j) = q \cdot x + (r + j)$, the rule is sound.
+\end{lemma}
 
-\paragraph{Linear and AddCheckConcrete}
-For the \textit{Linear} and \textit{AddCheckConcrete} agents we have the interaction rule:
-$$
-\mathit{Linear}(y, s, t) \bowtie \mathit{AddCheckConcrete}(\mathit{out}, k) \Rightarrow \mathit{Linear}(y, s, t + k) \sim \mathit{out} \\
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
-		& \llbracket \mathit{AddCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k + w \\
-		& \Rightarrow \mathit{out} = k + (s \cdot y + t)
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: } & \llbracket \mathit{Linear}(y, s, t + k) \sim \mathit{out} \rrbracket \\
-		& \Rightarrow \mathit{out} = s \cdot y + (t + k)
-	\end{aligned}
-\end{aligned}
-$$
-Since $k + (s \cdot y + t) = s \cdot y + (t + k)$, the rule is sound.
+\begin{lemma}
+	For the \textit{Concrete} and \textit{MulCheckLinear} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(j) \bowtie \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \mathit{Linear}(x, q \cdot j, r \cdot j) \sim \mathit{out} \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
+			& \llbracket \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot w \cdot x + r \cdot w \\
+			& \Rightarrow \mathit{out} = q \cdot j \cdot x + r \cdot j
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Linear}(x, q \cdot j, r \cdot j) \sim \mathit{out} \rrbracket \\
+			& \Rightarrow \mathit{out} = (q \cdot j) \cdot x + (r \cdot j)
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $q \cdot j \cdot x + r \cdot j = (q \cdot j) \cdot x + (r \cdot j)$, the rule is sound.
+\end{lemma}
 
-\paragraph{Linear and MulCheckConcrete}
-For the \textit{Linear} and \textit{MulCheckConcrete} agents we have the interaction rule:
-$$
-\mathit{Linear}(y, s, t) \bowtie \mathit{MulCheckConcrete}(\mathit{out}, k) \Rightarrow \mathit{Linear}(y, s \cdot k, t \cdot k) \sim \mathit{out} \\
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
-		& \llbracket \mathit{MulCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k \cdot w \\
-		& \Rightarrow \mathit{out} = k \cdot (s \cdot y + t)
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: } & \llbracket \mathit{Linear}(y, s \cdot k, t \cdot k) \sim \mathit{out} \rrbracket \\
-		& \Rightarrow \mathit{out} = (s \cdot k) \cdot y + (t \cdot k)
-	\end{aligned}
-\end{aligned}
-$$
-Since $k \cdot (s \cdot y + t) = (s \cdot k) \cdot y + (t \cdot k)$, the rule is sound.
+\begin{lemma}
+	For the \textit{Linear} and \textit{AddCheckConcrete} agents we have the interaction rule:
+	$$
+	\mathit{Linear}(y, s, t) \bowtie \mathit{AddCheckConcrete}(\mathit{out}, k) \Rightarrow \mathit{Linear}(y, s, t + k) \sim \mathit{out} \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
+			& \llbracket \mathit{AddCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k + w \\
+			& \Rightarrow \mathit{out} = k + (s \cdot y + t)
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Linear}(y, s, t + k) \sim \mathit{out} \rrbracket \\
+			& \Rightarrow \mathit{out} = s \cdot y + (t + k)
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $k + (s \cdot y + t) = s \cdot y + (t + k)$, the rule is sound.
+\end{lemma}
 
-\paragraph{Concrete and AddCheckConcrete}
-For the \textit{Concrete} and \textit{AddCheckConcrete} agents we have the interaction rule:
-$$
-\mathit{Concrete}(j) \bowtie \mathit{AddCheckConcrete}(\mathit{out}, k) \Rightarrow
-\begin{cases}
-	\mathit{out} \sim \mathit{Concrete}(k) & \text{if } j = 0 \\
-	\mathit{out} \sim \mathit{Concrete}(k + j) & \text{otherwise}
-\end{cases}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
-		& \llbracket \mathit{AddCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k + w \\
-		& \Rightarrow \mathit{out} = k + j
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: }
-		\begin{cases}
-			\llbracket \mathit{out} \sim \mathit{Concrete}(k) \rrbracket \Rightarrow \mathit{out} = k & \text{if } j = 0 \\
-			\llbracket \mathit{out} \sim \mathit{Concrete}(k + j) \rrbracket \Rightarrow \mathit{out} = k + j & \text{otherwise}
-		\end{cases}
+\begin{lemma}
+	For the \textit{Linear} and \textit{MulCheckConcrete} agents we have the interaction rule:
+	$$
+	\mathit{Linear}(y, s, t) \bowtie \mathit{MulCheckConcrete}(\mathit{out}, k) \Rightarrow \mathit{Linear}(y, s \cdot k, t \cdot k) \sim \mathit{out} \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
+			& \llbracket \mathit{MulCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k \cdot w \\
+			& \Rightarrow \mathit{out} = k \cdot (s \cdot y + t)
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Linear}(y, s \cdot k, t \cdot k) \sim \mathit{out} \rrbracket \\
+			& \Rightarrow \mathit{out} = (s \cdot k) \cdot y + (t \cdot k)
+		\end{aligned}
 	\end{aligned}
-\end{aligned}
-$$
-Since:
-$$
-\begin{cases}
-	k + 0 = k & \text{if } j = 0 \\
-	k + j = k + j & \text{otherwise}
-\end{cases}
-$$
-the rule is sound.
+	$$
+	Since $k \cdot (s \cdot y + t) = (s \cdot k) \cdot y + (t \cdot k)$, the rule is sound.
+\end{lemma}
 
-\paragraph{Concrete and MulCheckConcrete}
-For the \textit{Concrete} and \textit{MulCheckConcrete} agents we have the interaction rule:
-$$
-\mathit{Concrete}(j) \bowtie \mathit{MulCheckConcrete}(\mathit{out}, k) \Rightarrow
-\begin{cases}
-	\mathit{out} \sim \mathit{Concrete}(0) & \text{if } j = 0 \\
-	\mathit{out} \sim \mathit{Concrete}(k) & \text{if } j = 1 \\
-	\mathit{out} \sim \mathit{Concrete}(k \cdot j) & \text{otherwise}
-\end{cases}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
-		& \llbracket \mathit{MulCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k \cdot w \\
-		& \Rightarrow \mathit{out} = k \cdot j
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: }
-		\begin{cases}
-			\llbracket \mathit{out} \sim \mathit{Concrete}(0) \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } j = 0 \\
-			\llbracket \mathit{out} \sim \mathit{Concrete}(k) \rrbracket \Rightarrow \mathit{out} = k & \text{if } j = 1 \\
-			\llbracket \mathit{out} \sim \mathit{Concrete}(k \cdot j) \rrbracket \Rightarrow \mathit{out} = k \cdot j & \text{otherwise}
-		\end{cases}
+\begin{lemma}
+	For the \textit{Concrete} and \textit{AddCheckConcrete} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(j) \bowtie \mathit{AddCheckConcrete}(\mathit{out}, k) \Rightarrow
+	\begin{cases}
+		\mathit{out} \sim \mathit{Concrete}(k) & \text{if } j = 0 \\
+		\mathit{out} \sim \mathit{Concrete}(k + j) & \text{otherwise}
+	\end{cases}
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
+			& \llbracket \mathit{AddCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k + w \\
+			& \Rightarrow \mathit{out} = k + j
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+			\begin{cases}
+				\llbracket \mathit{out} \sim \mathit{Concrete}(k) \rrbracket \Rightarrow \mathit{out} = k & \text{if } j = 0 \\
+				\llbracket \mathit{out} \sim \mathit{Concrete}(k + j) \rrbracket \Rightarrow \mathit{out} = k + j & \text{otherwise}
+			\end{cases}
+		\end{aligned}
 	\end{aligned}
-\end{aligned}
-$$
-Since:
-$$
-\begin{cases}
-	k \cdot 0 = 0 & \text{if } j = 0 \\
-	k \cdot 1 = k & \text{if } j = 1 \\
-	k \cdot j = k \cdot j & \text{otherwise}
-\end{cases}
-$$
-the rule is sound.
+	$$
+	Since:
+	$$
+	\begin{cases}
+		k + 0 = k & \text{if } j = 0 \\
+		k + j = k + j & \text{otherwise}
+	\end{cases}
+	$$
+	the rule is sound.
+\end{lemma}
 
-\paragraph{Linear and ReLU}
-For the \textit{Linear} and \textit{ReLU} agents we have the interaction rule:
-$$
-\begin{aligned}
-	& \mathit{Linear}(x, q, r) \bowtie \mathit{ReLU}(\mathit{out}) \Rightarrow
-	\mathit{Linear}(x, q, r) \sim \mathit{Materialize}(\mathit{out}_x);
-	\mathit{out} \sim \mathit{Linear}(\mathit{TermReLU}(\mathit{out}_x), 1, 0)
-\end{aligned}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Linear}(q, x, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
-		& \llbracket \mathit{ReLU}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} = \max(0, w) \\
-		& \Rightarrow \mathit{out} = \max(0, q \cdot x + r)
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow w = q \cdot x + r \\
-		& \llbracket \mathit{Materialize}(\mathit{out}_x) \sim w \rrbracket \Rightarrow \mathit{out}_x = w \\
-		& \llbracket \mathit{Linear}(\mathit{TermReLU}(\mathit{out}_x), 1, 0) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = 1 \cdot \max(0, \mathit{out}_x) + 0 \\
-		& \Rightarrow \mathit{out} = 1 \cdot \max(0, q \cdot x + r) + 0
-	\end{aligned}
-\end{aligned}
-$$
-Since $\max(0, q \cdot x + r) = 1 \cdot \max(0, q \cdot x + r) + 0$ the rule is sound.
+\begin{lemma}
+	For the \textit{Concrete} and \textit{MulCheckConcrete} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(j) \bowtie \mathit{MulCheckConcrete}(\mathit{out}, k) \Rightarrow
+	\begin{cases}
+		\mathit{out} \sim \mathit{Concrete}(0) & \text{if } j = 0 \\
+		\mathit{out} \sim \mathit{Concrete}(k) & \text{if } j = 1 \\
+		\mathit{out} \sim \mathit{Concrete}(k \cdot j) & \text{otherwise}
+	\end{cases}
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
+			& \llbracket \mathit{MulCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k \cdot w \\
+			& \Rightarrow \mathit{out} = k \cdot j
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+			\begin{cases}
+				\llbracket \mathit{out} \sim \mathit{Concrete}(0) \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } j = 0 \\
+				\llbracket \mathit{out} \sim \mathit{Concrete}(k) \rrbracket \Rightarrow \mathit{out} = k & \text{if } j = 1 \\
+				\llbracket \mathit{out} \sim \mathit{Concrete}(k \cdot j) \rrbracket \Rightarrow \mathit{out} = k \cdot j & \text{otherwise}
+			\end{cases}
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since:
+	$$
+	\begin{cases}
+		k \cdot 0 = 0 & \text{if } j = 0 \\
+		k \cdot 1 = k & \text{if } j = 1 \\
+		k \cdot j = k \cdot j & \text{otherwise}
+	\end{cases}
+	$$
+	the rule is sound.
+\end{lemma}
 
-\paragraph{Concrete and ReLU}
-For the \textit{Concrete} and \textit{ReLU} agents we have the interaction rule:
-$$
-\mathit{Concrete}(k) \bowtie \mathit{ReLU}(\mathit{out}) \Rightarrow
-\begin{cases}
-	\mathit{out} \sim \mathit{Concrete}(k) & \text{if } k > 0 \\
-	\mathit{out} \sim \mathit{Concrete}(0) & \text{otherwise}
-\end{cases}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
-		& \llbracket \mathit{ReLU}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} = \max(0, w) \\
-		& \Rightarrow \mathit{out} = \max(0, k)
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: }
-		\begin{cases}
-			\llbracket \mathit{out} \sim \mathit{Concrete}(k) \rrbracket \Rightarrow \mathit{out} = k & \text{if } k > 0 \\
-			\llbracket \mathit{out} \sim \mathit{Concrete}(0) \rrbracket \Rightarrow \mathit{out} = 0 & \text{otherwise}
-		\end{cases}
+\begin{lemma}
+	For the \textit{Linear} and \textit{ReLU} agents we have the interaction rule:
+	$$
+	\begin{aligned}
+		\mathit{Linear}(x, q, r) \bowtie \mathit{ReLU}(\mathit{out}) \Rightarrow
+		& \mathit{Linear}(x, q, r) \sim \mathit{Materialize}(\mathit{out}_x); \\
+		& \mathit{out} \sim \mathit{Linear}(\mathit{TermReLU}(\mathit{out}_x), 1, 0)
+	\end{aligned}
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
+			& \llbracket \mathit{ReLU}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} = \max(0, w) \\
+			& \Rightarrow \mathit{out} = \max(0, q \cdot x + r)
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow w = q \cdot x + r \\
+			& \llbracket \mathit{Materialize}(\mathit{out}_x) \sim w \rrbracket \Rightarrow \mathit{out}_x = w \\
+			& \llbracket \mathit{Linear}(\mathit{TermReLU}(\mathit{out}_x), 1, 0) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = 1 \cdot \max(0, \mathit{out}_x) + 0 \\
+			& \Rightarrow \mathit{out} = 1 \cdot \max(0, q \cdot x + r) + 0
+		\end{aligned}
 	\end{aligned}
-\end{aligned}
-$$
-Since:
-$$
-\begin{cases}
-	\max(0, k) = k & \text{if } k > 0 \\
-	\max(0, k) = 0 & \text{otherwise}
-\end{cases}
-$$
-the rule is sound.
+	$$
+	Since $\max(0, q \cdot x + r) = 1 \cdot \max(0, q \cdot x + r) + 0$ the rule is sound.
+\end{lemma}
 
-\paragraph{Linear and Materialize}
-For the \textit{Linear} and \textit{Materialize} agents we have the interaction rule:
-$$
-\begin{aligned}
-	& \mathit{Linear}(x, q, r) \bowtie \mathit{Materialize}(\mathit{out}) \Rightarrow \\
-	& \quad \begin{cases}
-		\mathit{out} \sim \mathit{Concrete}(r); x \sim \mathit{Eraser} & \text{if } q = 0 \\
-		\mathit{out} \sim x & \text{if } q = 1, r = 0 \\
-		\mathit{out} \sim \mathit{TermAdd}(x, \mathit{Concrete}(r)) & \text{if } q = 1 \\
-		\mathit{out} \sim \mathit{TermMul}(\mathit{Concrete}(q), x) & \text{if } r = 0 \\
-		\mathit{out} \sim \mathit{TermAdd}(\mathit{TermMul}(\mathit{Concrete}(q), x), \mathit{Concrete}(r)) & \text{otherwise}
+\begin{lemma}
+	For the \textit{Concrete} and \textit{ReLU} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(k) \bowtie \mathit{ReLU}(\mathit{out}) \Rightarrow
+	\begin{cases}
+		\mathit{out} \sim \mathit{Concrete}(k) & \text{if } k > 0 \\
+		\mathit{out} \sim \mathit{Concrete}(0) & \text{otherwise}
+	\end{cases}
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
+			& \llbracket \mathit{ReLU}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} = \max(0, w) \\
+			& \Rightarrow \mathit{out} = \max(0, k)
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+			\begin{cases}
+				\llbracket \mathit{out} \sim \mathit{Concrete}(k) \rrbracket \Rightarrow \mathit{out} = k & \text{if } k > 0 \\
+				\llbracket \mathit{out} \sim \mathit{Concrete}(0) \rrbracket \Rightarrow \mathit{out} = 0 & \text{otherwise}
+			\end{cases}
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since:
+	$$
+	\begin{cases}
+		\max(0, k) = k & \text{if } k > 0 \\
+		\max(0, k) = 0 & \text{otherwise}
 	\end{cases}
-\end{aligned}
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
-		& \llbracket \mathit{Materialize}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} = w \\
-		& \Rightarrow \mathit{out} = q \cdot x + r
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: }
-		\begin{cases}
-			\llbracket \mathit{out} \sim \mathit{Concrete}(r); x \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = r & \text{if } q = 0 \\
-			\llbracket \mathit{out} \sim x \rrbracket \Rightarrow \mathit{out} = x & \text{if } q = 1, r = 0 \\
-			\llbracket \mathit{out} \sim \mathit{TermAdd}(x, \mathit{Concrete}(r)) \rrbracket \Rightarrow \mathit{out} = x + r & \text{if } q = 1 \\
-			\llbracket \mathit{out} \sim \mathit{TermMul}(\mathit{Concrete}(q), x) \rrbracket \Rightarrow \mathit{out} = q \cdot x & \text{if } r = 0 \\
-			\llbracket \mathit{out} \sim \mathit{TermAdd}(\mathit{TermMul}(\mathit{Concrete}(q), x), \mathit{Concrete}(r)) \rrbracket \Rightarrow \mathit{out} = q \cdot x + r & \text{otherwise} \\
+	$$
+	the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{Linear} and \textit{Materialize} agents we have the interaction rule:
+	$$
+	\begin{aligned}
+		& \mathit{Linear}(x, q, r) \bowtie \mathit{Materialize}(\mathit{out}) \Rightarrow \\
+		& \quad \begin{cases}
+			\mathit{out} \sim \mathit{TermConcrete}(r); x \sim \mathit{Eraser} & \text{if } q = 0 \\
+			\mathit{out} \sim x & \text{if } q = 1, r = 0 \\
+			\mathit{out} \sim \mathit{TermAdd}(x, \mathit{TermConcrete}(r)) & \text{if } q = 1 \\
+			\mathit{out} \sim \mathit{TermMul}(\mathit{TermConcrete}(q), x) & \text{if } r = 0 \\
+			\mathit{out} \sim \mathit{TermAdd}(\mathit{TermMul}(\mathit{TermConcrete}(q), x), \mathit{TermConcrete}(r)) & \text{otherwise}
 		\end{cases}
 	\end{aligned}
-\end{aligned}
-$$
-Since:
-$$
-\begin{cases}
-	0 \cdot x + r = r & \text{if } q = 0 \\
-	1 \cdot x + 0 = x & \text{if } q = 1, r = 0 \\
-	1 \cdot x + r = x + r & \text{if } q = 1 \\
-	q \cdot x + 0 = q	\cdot x & \text{if } r = 0 \\
-	q \cdot x + r = q \cdot x + r & \text{otherwise} \\
-\end{cases}
-$$
-the rule is sound.
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
+			& \llbracket \mathit{Materialize}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} = w \\
+			& \Rightarrow \mathit{out} = q \cdot x + r
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+			\begin{cases}
+				\llbracket \mathit{out} \sim \mathit{TermConcrete}(r); x \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = r & \text{if } q = 0 \\
+				\llbracket \mathit{out} \sim x \rrbracket \Rightarrow \mathit{out} = x & \text{if } q = 1, r = 0 \\
+				\llbracket \mathit{out} \sim \mathit{TermAdd}(x, \mathit{TermConcrete}(r)) \rrbracket \Rightarrow \mathit{out} = x + r & \text{if } q = 1 \\
+				\llbracket \mathit{out} \sim \mathit{TermMul}(\mathit{TermConcrete}(q), x) \rrbracket \Rightarrow \mathit{out} = q \cdot x & \text{if } r = 0 \\
+				\llbracket \mathit{out} \sim \mathit{TermAdd}(\mathit{TermMul}(\mathit{TermConcrete}(q), x), \mathit{TermConcrete}(r)) \rrbracket \Rightarrow \mathit{out} = q \cdot x + r & \text{otherwise} \\
+			\end{cases}
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since:
+	$$
+	\begin{cases}
+		0 \cdot x + r = r & \text{if } q = 0 \\
+		1 \cdot x + 0 = x & \text{if } q = 1, r = 0 \\
+		1 \cdot x + r = x + r & \text{if } q = 1 \\
+		q \cdot x + 0 = q	\cdot x & \text{if } r = 0 \\
+		q \cdot x + r = q \cdot x + r & \text{otherwise} \\
+	\end{cases}
+	$$
+	the rule is sound.
+\end{lemma}
 
-\paragraph{Concrete and Materialize}
-For the \textit{Concrete} and \textit{Materialize} agents we have the interaction rule:
-$$
-\mathit{Concrete}(k) \bowtie \mathit{Materialize}(\mathit{out}) \Rightarrow \mathit{Concrete}(k) \sim \mathit{out} \\
-$$
-We need to show that the LHS and RHS are semantically equivalent:
-$$
-\begin{aligned}
-	& \begin{aligned}
-		\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
-		& \llbracket \mathit{Materialize}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} =  w \\
-		& \Rightarrow \mathit{out} = k
-	\end{aligned} \\
-	& \begin{aligned}
-		\text{RHS: } & \llbracket \mathit{Concrete}(k) \sim \mathit{out} \rrbracket \\
-		& \Rightarrow \mathit{out} = k 
-	\end{aligned}
-\end{aligned}
-$$
-Since $k = k$, the rule is sound.
+\begin{lemma}
+	For the \textit{Concrete} and \textit{Materialize} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(k) \bowtie \mathit{Materialize}(\mathit{out}) \Rightarrow \mathit{TermConcrete}(k) \sim \mathit{out} \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
+			& \llbracket \mathit{Materialize}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} =  w \\
+			& \Rightarrow \mathit{out} = k
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{TermConcrete}(k) \sim \mathit{out} \rrbracket \\
+			& \Rightarrow \mathit{out} = k 
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $k = k$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{Linear} and \textit{Dup} agents we have the interaction rule:
+	$$
+	\mathit{Linear}(z, q, r) \bowtie \mathit{Dup}(x, y) \Rightarrow \mathit{Linear}(z_1, q, r) \sim x; \mathit{Linear}(z_2, q, r) \sim y; \mathit{Dup}(z_1, z_2) \sim z \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(z, q, r) \sim w \rrbracket \Rightarrow z\cdot q + r = w \\
+			& \llbracket \mathit{Dup}(x, y) \sim w \rrbracket \Rightarrow w = x = y \\
+			& \Rightarrow z \cdot q + r = x = y
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Linear}(z_1, q, r) \sim x \rrbracket \Rightarrow z_1 \cdot q + r = x \\
+			& \llbracket \mathit{Linear}(z_2, q, r) \sim y \rrbracket \Rightarrow z_2 \cdot q + r = y \\
+			& \llbracket \mathit{Dup}(z_1, z_2) \sim z \rrbracket \Rightarrow z = z_1 = z_2 \\
+			& \Rightarrow z \cdot q + r = x \land z \cdot q + r = y
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $z \cdot q + r = x = y \iff z \cdot q + r = x \land z \cdot q + r = y$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{Linear} and \textit{Eraser} agents we have the interaction rule:
+	$$
+	\mathit{Linear}(x, q, r) \bowtie \mathit{Eraser} \Rightarrow \mathit{Eraser} \sim x \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow x\cdot q + r = w \\
+			& \llbracket \mathit{Eraser} \sim w \rrbracket \Rightarrow w \in \mathbb{R} \\
+			& \Rightarrow x \cdot q + r \in \mathbb{R}
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Eraser} \sim x \rrbracket \\
+			& \Rightarrow x \in \mathbb{R}
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $x \cdot q + r \in \mathbb{R} \iff x \in \mathbb{R}$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{Concrete} and \textit{Dup} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(k) \bowtie \mathit{Dup}(x, y) \Rightarrow \mathit{Concrete}(k) \sim x; \mathit{Concrete}(k) \sim y \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
+			& \llbracket \mathit{Dup}(x, y) \sim w \rrbracket \Rightarrow w = x = y \\
+			& \Rightarrow k = x = y
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Concrete}(k) \sim x \rrbracket \Rightarrow k = x \\
+			& \llbracket \mathit{Concrete}(k) \sim y \rrbracket \Rightarrow k = y \\
+			& \Rightarrow k = x \land k = y
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $k = x = y \iff k = x \land k = y$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{Concrete} and \textit{Eraser} agents we have the interaction rule:
+	$$
+	\mathit{Concrete}(k) \bowtie \mathit{Eraser} \Rightarrow  \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
+			& \llbracket \mathit{Eraser} \sim w \rrbracket \Rightarrow w \in \mathbb{R} \\
+			& \Rightarrow k \in \mathbb{R} 
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $k \in \mathbb{R}$ is not violated by RHS, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermAdd} and \textit{Dup} agents we have the interaction rule:
+	$$
+	\begin{aligned}
+		\mathit{TermAdd}(a, b) \bowtie \mathit{Dup}(x, y) \Rightarrow & \mathit{TermAdd}(a_1, b_1) \sim x; \mathit{TermAdd}(a_2, b_2) \sim y; \\
+		& \mathit{Dup}(a_1, a_2) \sim a; \mathit{Dup}(b_1, b_2) \sim b\\
+	\end{aligned}
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermAdd}(a, b) \sim w \rrbracket \Rightarrow a + b = w \\
+			& \llbracket \mathit{Dup}(x, y) \sim w \rrbracket \Rightarrow w = x = y \\
+			& \Rightarrow a + b = x = y
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{TermAdd}(a_1, b_1) \sim x \rrbracket \Rightarrow a_1 + b_1 = x \\
+			& \llbracket \mathit{TermAdd}(a_2, b_2) \sim y \rrbracket \Rightarrow a_2 + b_2 = y \\
+			& \llbracket \mathit{Dup}(a_1, a_2) \sim a \rrbracket \Rightarrow a = a_1 = a_2 \\
+			& \llbracket \mathit{Dup}(b_1, b_2) \sim b \rrbracket \Rightarrow b = b_1 = b_2 \\
+			& \Rightarrow a + b = x \land a + b = y
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $a + b = x = y \iff a + b = x \land a + b = y$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermAdd} and \textit{Eraser} agents we have the interaction rule:
+	$$
+	\mathit{TermAdd}(a, b) \bowtie \mathit{Eraser} \Rightarrow \mathit{Eraser} \sim a; \mathit{Eraser} \sim b \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermAdd}(a, b) \sim w \rrbracket \Rightarrow a + b = w \\
+			& \llbracket \mathit{Eraser} \sim w \rrbracket \Rightarrow w \in \mathbb{R} \\
+			& \Rightarrow a + b \in \mathbb{R}
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Eraser} \sim a \rrbracket \Rightarrow a \in \mathbb{R} \\
+			& \llbracket \mathit{Eraser} \sim b \rrbracket \Rightarrow b \in \mathbb{R} \\
+			& \Rightarrow a \in \mathbb{R} \land b \in \mathbb{R}
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $a + b \in \mathbb{R} \iff a \in \mathbb{R} \land b \in \mathbb{R}$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermMul} and \textit{Dup} agents we have the interaction rule:
+	$$
+	\begin{aligned}
+		\mathit{TermMul}(a, b) \bowtie \mathit{Dup}(x, y) \Rightarrow & \mathit{TermMul}(a_1, b_1) \sim x; \mathit{TermMul}(a_2, b_2) \sim y; \\
+		& \mathit{Dup}(a_1, a_2) \sim a; \mathit{Dup}(b_1, b_2) \sim b\\
+	\end{aligned}
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermMul}(a, b) \sim w \rrbracket \Rightarrow a \cdot b = w \\
+			& \llbracket \mathit{Dup}(x, y) \sim w \rrbracket \Rightarrow w = x = y \\
+			& \Rightarrow a \cdot b = x = y
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{TermMul}(a_1, b_1) \sim x \rrbracket \Rightarrow a_1 \cdot b_1 = x \\
+			& \llbracket \mathit{TermMul}(a_2, b_2) \sim y \rrbracket \Rightarrow a_2 \cdot b_2 = y \\
+			& \llbracket \mathit{Dup}(a_1, a_2) \sim a \rrbracket \Rightarrow a = a_1 = a_2 \\
+			& \llbracket \mathit{Dup}(b_1, b_2) \sim b \rrbracket \Rightarrow b = b_1 = b_2 \\
+			& \Rightarrow a \cdot b = x \land a \cdot b = y
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $a \cdot b = x = y \iff a \cdot b = x \land a \cdot b = y$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermMul} and \textit{Eraser} agents we have the interaction rule:
+	$$
+	\mathit{TermMul}(a, b) \bowtie \mathit{Eraser} \Rightarrow \mathit{Eraser} \sim a; \mathit{Eraser} \sim b \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermMul}(a, b) \sim w \rrbracket \Rightarrow a \cdot b = w \\
+			& \llbracket \mathit{Eraser} \sim w \rrbracket \Rightarrow w \in \mathbb{R} \\
+			& \Rightarrow a \cdot b \in \mathbb{R}
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Eraser} \sim a \rrbracket \Rightarrow a \in \mathbb{R} \\
+			& \llbracket \mathit{Eraser} \sim b \rrbracket \Rightarrow b \in \mathbb{R} \\
+			& \Rightarrow a \in \mathbb{R} \land b \in \mathbb{R}
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $a \cdot b \in \mathbb{R} \iff a \in \mathbb{R} \land b \in \mathbb{R}$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermReLU} and \textit{Dup} agents we have the interaction rule:
+	$$
+	\mathit{TermReLU}(z) \bowtie \mathit{Dup}(x, y) \Rightarrow \mathit{TermReLU}(z_1) \sim x; \mathit{TermReLU}(z_2) \sim y; \mathit{Dup}(z_1, z_2) \sim z \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermReLU}(z) \sim w \rrbracket \Rightarrow \max(0, z) = w \\
+			& \llbracket \mathit{Dup}(x, y) \sim w \rrbracket \Rightarrow w = x = y \\
+			& \Rightarrow \max(0, z) = x = y
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{TermReLU}(z_1) \sim x \rrbracket \Rightarrow \max(0, z_1) = x \\
+			& \llbracket \mathit{TermReLU}(z_2) \sim y \rrbracket \Rightarrow \max(0, z_2) = y \\
+			& \llbracket \mathit{Dup}(z_1, z_2) \sim z \rrbracket \Rightarrow z = z_1 = z_2 \\
+			& \Rightarrow \max(0, z) = x \land \max(0, z) = y
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $\max(0, z) = x = y \iff \max(0, z) = x \land \max(0, z) = y$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermReLU} and \textit{Eraser} agents we have the interaction rule:
+	$$
+	\mathit{TermReLU}(x) \bowtie \mathit{Eraser} \Rightarrow \mathit{Eraser} \sim x \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermReLU}(x) \sim w \rrbracket \Rightarrow \max(0, x) = w \\
+			& \llbracket \mathit{Eraser} \sim w \rrbracket \Rightarrow w \in \mathbb{R} \\
+			& \Rightarrow \max(0, x) \in \mathbb{R}
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{Eraser} \sim x \rrbracket \\
+			& \Rightarrow x \in \mathbb{R}
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $\max(0, x) \in \mathbb{R} \iff x \in \mathbb{R}$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermConcrete} and \textit{Dup} agents we have the interaction rule:
+	$$
+	\mathit{TermConcrete}(k) \bowtie \mathit{Dup}(x, y) \Rightarrow \mathit{TermConcrete}(k) \sim x; \mathit{TermConcrete}(k) \sim y \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermConcrete}(k) \sim w \rrbracket \Rightarrow k = w \\
+			& \llbracket \mathit{Dup}(x, y) \sim w \rrbracket \Rightarrow w = x = y \\
+			& \Rightarrow k = x = y
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{TermConcrete}(k) \sim x \rrbracket \Rightarrow k = x \\
+			& \llbracket \mathit{TermConcrete}(k) \sim y \rrbracket \Rightarrow k = y \\
+			& \Rightarrow k = x \land k = y
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $k = x = y \iff k = x \land k = y$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermConcrete} and \textit{Eraser} agents we have the interaction rule:
+	$$
+	\mathit{TermConcrete}(k) \bowtie \mathit{Eraser} \Rightarrow  \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermConcrete}(k) \sim w \rrbracket \Rightarrow k = w \\
+			& \llbracket \mathit{Eraser} \sim w \rrbracket \Rightarrow w \in \mathbb{R} \\
+			& \Rightarrow k \in \mathbb{R} 
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $k \in \mathbb{R}$ is not violated by RHS, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermSymbolic} and \textit{Dup} agents we have the interaction rule:
+	$$
+	\mathit{TermSymbolic}(id) \bowtie \mathit{Dup}(x, y) \Rightarrow \mathit{TermSymbolic}(id) \sim x; \mathit{TermSymbolic}(id) \sim y \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermSymbolic}(id) \sim w \rrbracket \Rightarrow x_{id} = w \\
+			& \llbracket \mathit{Dup}(x, y) \sim w \rrbracket \Rightarrow w = x = y \\
+			& \Rightarrow x_{id} = x = y
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: } & \llbracket \mathit{TermSymbolic}(id) \sim x \rrbracket \Rightarrow x_{id} = x \\
+			& \llbracket \mathit{TermSymbolic}(id) \sim y \rrbracket \Rightarrow x_{id} = y \\
+			& \Rightarrow x_{id} = x \land x_{id} = y
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $x_{id} = x = y \iff x_{id} = x \land x_{id} = y$, the rule is sound.
+\end{lemma}
+
+\begin{lemma}
+	For the \textit{TermSymbolic} and \textit{Eraser} agents we have the interaction rule:
+	$$
+	\mathit{TermSymbolic}(id) \bowtie \mathit{Eraser} \Rightarrow  \\
+	$$
+	We need to show that the LHS and RHS are semantically equivalent:
+	$$
+	\begin{aligned}
+		& \begin{aligned}
+			\text{LHS: } & \llbracket \mathit{TermSymbolic}(id) \sim w \rrbracket \Rightarrow x_{id} = w \\
+			& \llbracket \mathit{Eraser} \sim w \rrbracket \Rightarrow w \in \mathbb{R} \\
+			& \Rightarrow x_{id} \in \mathbb{R} 
+		\end{aligned} \\
+		& \begin{aligned}
+			\text{RHS: }
+		\end{aligned}
+	\end{aligned}
+	$$
+	Since $x_{id} \in \mathbb{R}$ is not violated by RHS, the rule is sound.
+\end{lemma}
diff --git a/chapters/core/soundness-proof/04-soundness-of-reduction.tex b/chapters/core/soundness-proof/04-soundness-of-reduction.tex
index b0a7906..d0cec53 100644
--- a/chapters/core/soundness-proof/04-soundness-of-reduction.tex
+++ b/chapters/core/soundness-proof/04-soundness-of-reduction.tex
@@ -1,28 +1,117 @@
 \subsection{Soundness of Reduction}
 \label{sec:soundness-of-reduction}
 
-Let $\text{IN}_0$ be the IN translated from a neural network $\text{NN}$. Let $\text{IN}_n$ be
-the IN after $n$ reduction steps. Then we need to prove:
-$$
-\forall n \in \mathbb{N} \quad \llbracket \text{IN}_n \rrbracket = \llbracket \text{NN} \rrbracket
-$$
+\begin{lemma}
+	A valid IN satisfies the following properties:
+	\begin{itemize}
+		\item \textbf{DAG}: the net forms a DAG where the roots are the free wires representing the network
+			outputs.
+		\item \textbf{Orientation}: carrier agents always have their principal ports oriented toward the outputs,
+			while operator and intermediate agents always have their principal ports oriented toward the
+			inputs. No interaction rule introduces carriers facing the input nor operators or
+			intermediates facing the output.
+		\item \textbf{Restricted Interaction}: the net is constructed such that active pairs only occur between a
+			carrier agent and an operator/intermediate agent. Because of \textbf{Orientation}, neither
+			carrier nor operator agents interact with agents of the same type. Terminal agents do not
+			interact with computational operators because they are wrapped in a \textit{Linear} agent.
+	\end{itemize}
+\end{lemma}
 
-\paragraph{Base case: $n = 0$} By \textbf{\Cref{sec:soundness-of-translation}}, the initial
-$\text{IN}_0$ is constructed such that its semantics $\llbracket \text{IN}_0 \rrbracket$ exactly
-match the mathematical definition of the ONNX operators in $\text{NN}$.
+\begin{theorem}
+	Let $\text{IN}_0$ be the valid IN translated from a neural network $\text{NN}$. Let $\text{IN}_n$ be
+	the IN after $n$ reduction steps, then:
+	\begin{equation}
+	\forall n \in \mathbb{N} \quad \llbracket \text{IN}_n \rrbracket = \llbracket \text{NN} \rrbracket
+	\end{equation}
+\end{theorem}
 
-\paragraph{Induction step: $n \to n + 1$} Assume $\llbracket \text{IN}_n \rrbracket = \llbracket \text{NN} \rrbracket$.
-If $\text{IN}_n$ is in normal form, the proof is complete. Otherwise, there exists an active pair
-$A \bowtie B$ that reduces $\text{IN}_n$ to $\text{IN}_{n+1}$. By \textbf{\Cref{sec:soundness-of-interaction-rules}},
-the mathematical definition is preserved after any reduction step, it follows that:
-$$
-\llbracket \text{IN}_{n+1} \rrbracket = \llbracket \text{IN}_{n} \rrbracket
-$$
-By the inductive hypothesis:
-$$
-\llbracket \text{IN}_{n+1} \rrbracket = \llbracket \text{NN} \rrbracket
-$$
+\begin{proof}
+	We will proceed by induction on the number $n$ of reduction steps:
+	\paragraph{Base case: $n = 0$} By \textbf{\Cref{sec:soundness-of-translation}}, the initial
+	$\text{IN}_0$ is constructed such that its semantics $\llbracket \text{IN}_0 \rrbracket$ exactly
+	match the mathematical definition of the ONNX operators in $\text{NN}$, it follows that:
+	\begin{equation}
+	\llbracket \text{IN}_{0} \rrbracket = \llbracket \text{NN} \rrbracket
+	\end{equation}
 
-By the principle of mathematical induction, $\text{IN}_n$ remains semantically equivalent to the
-original $\text{NN}$ at every step of the reduction process. Additionally, since IN are confluent,
-the reduced mathematical expression is unique regardless of order in which rules are applied.
+	\paragraph{Induction step: $n \to n + 1$} Assume $\llbracket \text{IN}_n \rrbracket = \llbracket \text{NN} \rrbracket$.
+	If $\text{IN}_n$ is in normal form, the proof is complete. Otherwise, there exists an active pair
+	$A \bowtie B$ that reduces $\text{IN}_n$ to $\text{IN}_{n+1}$. By \textbf{\Cref{sec:soundness-of-interaction-rules}},
+	the mathematical definition is preserved after any reduction step, it follows that:
+	\begin{equation}
+	\llbracket \text{IN}_{n+1} \rrbracket = \llbracket \text{IN}_{n} \rrbracket
+	\end{equation}
+	By the inductive hypothesis:
+	\begin{equation}
+	\llbracket \text{IN}_{n+1} \rrbracket = \llbracket \text{NN} \rrbracket
+	\end{equation}
+	By the principle of mathematical induction, $\text{IN}_n$ remains semantically equivalent to the
+	original $\text{NN}$ at every step of the reduction process.
+\end{proof}
+
+\begin{theorem}
+	For any valid $\text{IN}_0$ translated from a neural network $\text{NN}$, the reduction process
+	$\text{IN}_0 \to \text{IN}_1 \to \dots \to \text{IN}_n$ reaches a unique normal form, an IN with no active pair, in a
+	finite number of steps $n$.
+\end{theorem}
+
+\begin{proof}
+	We define a potential function $\Phi$. We need to show that:
+	\begin{equation}
+	\Phi(\text{IN}_n) > \Phi(\text{IN}_{n+1}) \quad \forall n \in \mathbb{N}
+	\end{equation}
+	The potential function is defined as:
+	\begin{equation}
+	\Phi(\text{IN}_n) = \sum_{a \in \text{Agents}(\text{IN}_n)} \begin{cases}
+		4 + 3^{D-d(a)} & \text{ if }a\text{ is of type computational operator} \\
+		3 + 3^{D-d(a)} & \text{ if }a\text{ is of type intermediate} \\
+		1 + 3^{D-d(a)} & \text{ if }a\text{ is a Materialize agent} \\
+		3^{D-d(a)} & \text{ if }a\text{ is of type structural operator} \\
+		0 & \text{ if }a\text{ is a Carrier, Terminal}
+	\end{cases}
+	\end{equation}
+	where $d(a)$ is the distance of the agent $a$ from the nearest output wire and $D$ the maximum depth
+	of $\text{IN}_0$,
+	We observe that every reduction step $n \to n+1$ strictly reduces $\Phi$:
+	\begin{itemize}
+		\item \textbf{Carrier with Computational Operator}: If a carrier $c$ interacts with an operator agent $o$, it
+			either gets pruned into a new carrier agent closer to the root:
+			\begin{equation}
+			\Phi(\text{IN}_n) = 4 + 3^{D-d(c)} > 0 = \Phi(\text{IN}_{n+1})
+			\end{equation}
+			or absorbed into an intermediate agent:
+			\begin{equation}
+			\Phi(\text{IN}_n) = 4 + 3^{D-d(c)} > 3 + 3^{D-d(c)} = \Phi(\text{IN}_{n+1})
+			\end{equation}
+		\item \textbf{Carrier with Intermediate}: If a carrier $c$ interacts with an intermediate agent $i$, it
+			either gets pruned into a new carrier agent closer to the root:
+			\begin{equation}
+			\Phi(\text{IN}_n) = 3 + 3^{D-d(i)} > 0 = \Phi(\text{IN}_{n+1})
+			\end{equation}
+			or temporary Linear agents are wired with Materialize agents, which are then forced to
+			interact to produce terminal agents, then wrapped into a new Linear agent:
+			\begin{equation}
+			\Phi(\text{IN}_n) = 3 + 3^{D-d(i)} > 1 + 3^{D-d(i)-2} + 1 + 3^{D-d(i)-2} = 2 + 2 \cdot 3^{D-d(i)-2} = \Phi(\text{IN}_{n+1})
+			\end{equation}
+			If the carrier $c$ interacts with a Materialize agent $i$:
+			\begin{equation}
+			\Phi(\text{IN}_n) = 1 + 3^{D-d(i)} > 0 = \Phi(\text{IN}_{n+1})
+			\end{equation}
+		\item \textbf{Carrier with Structural Operator}: If a carrier $c$ is duplicated with the \textit{Dup} or \textit{Eraser}
+			agent $a$, it traverses the agent structure:
+			\begin{equation}
+			\Phi(\text{IN}_n) = 3^{D-d(a)} > 3^{D-d(a)-1} = \Phi(\text{IN}_{n+1})
+			\end{equation}
+			if the Dup or Eraser agent interacts with a TermAdd or TermMul, two new agents are created at
+			each auxillary port:
+			\begin{equation}
+			\Phi(\text{IN}_n) = 3^{D-d(a)} > 2 * 3^{D-d(a)-1} = \Phi(\text{IN}_{n+1})
+			\end{equation}
+	\end{itemize}
+	Since $\Phi$ is a non-negative strictly decreasing function, the reduction
+	process must terminate in a finite number of steps $n$.
+
+	Additionally, since each active pair has exactly one applicable rule the reduction is
+	deterministic. Strong confluence follows immediately, meaning that the normal form is unique
+	regardless of order in which rules are applied.
+\end{proof}