\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.

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

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

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

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

\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.

\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.

\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.

\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.

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

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

\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.

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

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

\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.
