\subsection{Soundness of Interaction Rules}
\label{sec:soundness-of-interaction-rules}

\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}

\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:
	$$
	\begin{aligned}
		& \begin{aligned}
			\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
			& \llbracket \mathit{Mul}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w \cdot b \\
			& \Rightarrow \mathit{out} = (q \cdot x + r) \cdot b
		\end{aligned} \\
		& \begin{aligned}
			\text{RHS: } & \llbracket \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \sim b \rrbracket \\
			& \Rightarrow \mathit{out} = q \cdot b \cdot x + r \cdot b
		\end{aligned}
	\end{aligned}
	$$
	Since $(q \cdot x + r) \cdot b = q \cdot b \cdot x + r \cdot b$, the rule is sound.
\end{lemma}

\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:
	$$
	\begin{aligned}
		& \begin{aligned}
			\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
			& \llbracket \mathit{Add}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w + b \\
			& \Rightarrow \mathit{out} = k + b
		\end{aligned} \\
		& \begin{aligned}
			\text{RHS: }
			\begin{cases}
				\llbracket \mathit{out} \sim b \rrbracket \Rightarrow \mathit{out} = b & \text{if } k = 0 \\
				\llbracket \mathit{AddCheckConcrete}(\mathit{out}, k) \sim b \rrbracket \Rightarrow \mathit{out} = k + b & \text{otherwise}
			\end{cases}
		\end{aligned}
	\end{aligned}
	$$
	Since:
	$$
	\begin{cases}
		0 + b = b & \text{if } k = 0 \\
		k + b = k + b & \text{otherwise}
	\end{cases}
	$$
	the rule is sound.
\end{lemma}

\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:
	$$
	\begin{aligned}
		& \begin{aligned}
			\text{LHS: } & \llbracket \mathit{Concrete}(k) \sim w \rrbracket \Rightarrow k = w \\
			& \llbracket \mathit{Mul}(\mathit{out}, b) \sim w \rrbracket \Rightarrow \mathit{out} = w \cdot b \\
			& \Rightarrow \mathit{out} = k \cdot b
		\end{aligned} \\
		& \begin{aligned}
			\text{RHS: }
			\begin{cases}
				\llbracket \mathit{out} \sim \mathit{Concrete}(0); b \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } k = 0 \\
				\llbracket \mathit{out} \sim b \rrbracket \Rightarrow \mathit{out} = b & \text{if } k = 1 \\
				\llbracket \mathit{MulCheckConcrete}(\mathit{out}, k) \sim b \rrbracket \Rightarrow \mathit{out} = k \cdot b & \text{otherwise}
			\end{cases}
		\end{aligned}
	\end{aligned}
	$$
	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.
\end{lemma}

\begin{lemma}
	For the \textit{Linear} and \textit{AddCheckLinear} agents we have the interaction rule:
	$$
	\begin{aligned}
		& \mathit{Linear}(y, s, t) \bowtie \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \\
		& \quad \begin{cases}
			\mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} & \text{if } q,r,s,t = 0 \\
			\mathit{out} \sim  \mathit{Linear}(x, q, r); y \sim \mathit{Eraser} & \text{if } s,t = 0 \\
			\mathit{out} \sim  \mathit{Linear}(y, s, t); x \sim \mathit{Eraser} & \text{if } q,r = 0 \\
			\begin{aligned}
				& \mathit{Linear}(x, q, r) \sim \mathit{Materialize}(\mathit{out}_x); \\
				& \mathit{Linear}(y, s, t) \sim \mathit{Materialize}(\mathit{out}_y); \\
				& \mathit{out} \sim \mathit{Linear}(\mathit{TermAdd}(\mathit{out}_x, \mathit{out}_y), 1, 0)
			\end{aligned}
			& \text{otherwise}
		\end{cases}
	\end{aligned}
	$$
	We need to show that the LHS and RHS are semantically equivalent:
	$$
	\begin{aligned}
		& \begin{aligned}
			\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
			& \llbracket \mathit{AddCheckLinear}(\mathit{out}, x, q, r) \sim w \rrbracket \Rightarrow \mathit{out} = q \cdot x + (r + w) \\
			& \Rightarrow \mathit{out} = q \cdot x + (r + s \cdot y + t)
		\end{aligned} \\
		& \begin{aligned}
			\text{RHS: }
			\begin{cases}
				\llbracket \mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = 0 & \text{if } q,r,s,t = 0 \\
				\llbracket \mathit{out} \sim  \mathit{Linear}(x, q, r); y \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = q \cdot x + r & \text{if } s,t = 0 \\
				\llbracket \mathit{out} \sim  \mathit{Linear}(y, s, t); x \sim \mathit{Eraser} \rrbracket \Rightarrow \mathit{out} = s \cdot y + t & \text{if } q,r = 0 \\
				\begin{aligned}
					& \llbracket \mathit{Linear}(x, q, r) \sim w_1 \rrbracket \Rightarrow w_1 = q \cdot x + r \\
					& \llbracket \mathit{Materialize}(\mathit{out}_x) \sim w_1 \rrbracket \Rightarrow \mathit{out}_x = w_1 \\
					& \llbracket \mathit{Linear}(y, s, t) \sim w_2 \rrbracket \Rightarrow w_2 = s \cdot y + t \\
					& \llbracket \mathit{Materialize}(\mathit{out}_y) \sim w_2 \rrbracket \Rightarrow \mathit{out}_y = w_2 \\
					& \llbracket \mathit{Linear}(\mathit{TermAdd}(\mathit{out}_x, \mathit{out}_y), 1, 0) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = 1 \cdot (\mathit{out}_x + \mathit{out}_y) + 0
				\end{aligned}\\
				\Rightarrow \mathit{out} = 1 \cdot (q \cdot x + r + s \cdot y + t) + 0 & \text{otherwise}
			\end{cases}
		\end{aligned}
	\end{aligned}
	$$
	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.
\end{lemma}

\begin{lemma}
	For the \textit{Linear} and \textit{MulCheckLinear} agents we have the interaction rule:
	$$
	\begin{aligned}
		& \mathit{Linear}(y, s, t) \bowtie \mathit{MulCheckLinear}(\mathit{out}, x, q, r) \Rightarrow \\
		& \quad \begin{cases}
			\mathit{out} \sim \mathit{Concrete}(0); x \sim \mathit{Eraser}; y \sim \mathit{Eraser} & \text{if } q,r = 0 \lor s,t = 0 \\
			\begin{aligned}
				& \mathit{Linear}(x, q, r) \sim \mathit{Materialize}(\mathit{out}_x); \\
				& \mathit{Linear}(y, s, t) \sim \mathit{Materialize}(\mathit{out}_y); \\
				& \mathit{out} \sim \mathit{Linear}(\mathit{TermMul}(\mathit{out}_x, \mathit{out}_y), 1, 0)
			\end{aligned}
			& \text{otherwise}
		\end{cases}
	\end{aligned}
	$$
	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}

\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}

\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}

\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}

\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:
	$$
	\begin{aligned}
		& \begin{aligned}
			\text{LHS: } & \llbracket \mathit{Linear}(y, s, t) \sim w \rrbracket \Rightarrow s \cdot y + t = w \\
			& \llbracket \mathit{MulCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k \cdot w \\
			& \Rightarrow \mathit{out} = k \cdot (s \cdot y + t)
		\end{aligned} \\
		& \begin{aligned}
			\text{RHS: } & \llbracket \mathit{Linear}(y, s \cdot k, t \cdot k) \sim \mathit{out} \rrbracket \\
			& \Rightarrow \mathit{out} = (s \cdot k) \cdot y + (t \cdot k)
		\end{aligned}
	\end{aligned}
	$$
	Since $k \cdot (s \cdot y + t) = (s \cdot k) \cdot y + (t \cdot k)$, the rule is sound.
\end{lemma}

\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:
	$$
	\begin{aligned}
		& \begin{aligned}
			\text{LHS: } & \llbracket \mathit{Concrete}(j) \sim w \rrbracket \Rightarrow j = w \\
			& \llbracket \mathit{AddCheckConcrete}(\mathit{out}, k) \sim w \rrbracket \Rightarrow \mathit{out} = k + w \\
			& \Rightarrow \mathit{out} = k + j
		\end{aligned} \\
		& \begin{aligned}
			\text{RHS: }
			\begin{cases}
				\llbracket \mathit{out} \sim \mathit{Concrete}(k) \rrbracket \Rightarrow \mathit{out} = k & \text{if } j = 0 \\
				\llbracket \mathit{out} \sim \mathit{Concrete}(k + j) \rrbracket \Rightarrow \mathit{out} = k + j & \text{otherwise}
			\end{cases}
		\end{aligned}
	\end{aligned}
	$$
	Since:
	$$
	\begin{cases}
		k + 0 = k & \text{if } j = 0 \\
		k + j = k + j & \text{otherwise}
	\end{cases}
	$$
	the rule is sound.
\end{lemma}

\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}

\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:
	$$
	\begin{aligned}
		& \begin{aligned}
			\text{LHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow q \cdot x + r = w \\
			& \llbracket \mathit{ReLU}(\mathit{out}) \sim w \rrbracket \Rightarrow \mathit{out} = \max(0, w) \\
			& \Rightarrow \mathit{out} = \max(0, q \cdot x + r)
		\end{aligned} \\
		& \begin{aligned}
			\text{RHS: } & \llbracket \mathit{Linear}(x, q, r) \sim w \rrbracket \Rightarrow w = q \cdot x + r \\
			& \llbracket \mathit{Materialize}(\mathit{out}_x) \sim w \rrbracket \Rightarrow \mathit{out}_x = w \\
			& \llbracket \mathit{Linear}(\mathit{TermReLU}(\mathit{out}_x), 1, 0) \sim \mathit{out} \rrbracket \Rightarrow \mathit{out} = 1 \cdot \max(0, \mathit{out}_x) + 0 \\
			& \Rightarrow \mathit{out} = 1 \cdot \max(0, q \cdot x + r) + 0
		\end{aligned}
	\end{aligned}
	$$
	Since $\max(0, q \cdot x + r) = 1 \cdot \max(0, q \cdot x + r) + 0$ the rule is sound.
\end{lemma}

\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:
	$$
	\begin{cases}
		\max(0, k) = k & \text{if } k > 0 \\
		\max(0, k) = 0 & \text{otherwise}
	\end{cases}
	$$
	the rule is sound.
\end{lemma}

\begin{lemma}
	For the \textit{Linear} and \textit{Materialize} agents we have the interaction rule:
	$$
	\begin{aligned}
		& \mathit{Linear}(x, q, r) \bowtie \mathit{Materialize}(\mathit{out}) \Rightarrow \\
		& \quad \begin{cases}
			\mathit{out} \sim \mathit{TermConcrete}(r); x \sim \mathit{Eraser} & \text{if } q = 0 \\
			\mathit{out} \sim x & \text{if } q = 1, r = 0 \\
			\mathit{out} \sim \mathit{TermAdd}(x, \mathit{TermConcrete}(r)) & \text{if } q = 1 \\
			\mathit{out} \sim \mathit{TermMul}(\mathit{TermConcrete}(q), x) & \text{if } r = 0 \\
			\mathit{out} \sim \mathit{TermAdd}(\mathit{TermMul}(\mathit{TermConcrete}(q), x), \mathit{TermConcrete}(r)) & \text{otherwise}
		\end{cases}
	\end{aligned}
	$$
	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}

\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:
	$$
	\begin{aligned}
		\mathit{Linear}(z, q, r) \bowtie \mathit{Dup}(x, y) \Rightarrow
		& \mathit{Linear}(z_1, q, r) \sim x; \mathit{Linear}(z_2, q, r) \sim y;	\\
		& \mathit{Dup}(z_1, z_2) \sim z
	\end{aligned}
	$$
	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:
	$$
	\begin{aligned}
		\mathit{TermReLU}(z) \bowtie \mathit{Dup}(x, y) \Rightarrow
		& \mathit{TermReLU}(z_1) \sim x; \\
		&\mathit{TermReLU}(z_2) \sim y; \\
		& \mathit{Dup}(z_1, z_2) \sim z
	\end{aligned}
	$$
	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:
	$$
	\begin{aligned}
		\mathit{TermConcrete}(k) \bowtie \mathit{Dup}(x, y) \Rightarrow
		& \mathit{TermConcrete}(k) \sim x; \\
		& \mathit{TermConcrete}(k) \sim y
	\end{aligned}
	$$
	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:
	$$
	\begin{aligned}
		\mathit{TermSymbolic}(id) \bowtie \mathit{Dup}(x, y) \Rightarrow
		& \mathit{TermSymbolic}(id) \sim x; \\
		& \mathit{TermSymbolic}(id) \sim y
	\end{aligned}
	$$
	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}
