A. Proof of Theorem 4.1

Theorem A.1   Let $(P,\leq)$ be any finite partial order. Let $(2^{\mathbb{N}}, \subseteq)$ be the lattice of subsets of $\mathbb{N}$ with the set inclusion ordering. Then there exists a mapping $\phi : P \rightarrow 2^{\mathbb{N}}$, such that $\forall x,y \in P, x \leq y \iff \phi(x) \subseteq \phi(y)$ and $\phi(x)$ is a finite subset of $\mathbb{N}$ for all $x \in P$.

Proof : We prove the theorem by induction on $\vert P\vert$.
Base Case : Let $\vert P\vert$ = 1. Then the claim trivially holds.
Induction Hypothesis : Let the claim hold for all $P$ such that $\vert P\vert \leq k$.
Induction Step : $\vert P\vert = k+1$.
Let $(P,\leq)$ be a partial order such that $\vert P\vert = k+1$. Since $P$ is finite, $P$ has a minimal element, say $a$. Consider the partial order $(P\setminus \{a\},\leq)$. Clearly this is a partial order and $\vert P\setminus \{a\}\vert = k$. Hence by induction hypothesis, there exists $\phi : P\setminus \{a\} \rightarrow 2^{\mathbb{N}}$, such that $\forall x,y \in P\setminus \{a\} , x \leq y \iff \phi(x) \subseteq \phi(y)$ and $\phi(x)$ is a finite subset of $\mathbb{N}$ for all $x \in P\setminus \{a\}$. Let $n = \max_i\{i\in \cup_{x \in P\setminus \{a\}\}} \phi(x) \}$. Define $\phi' : P \rightarrow 2^{\mathbb{N}}$ as follows.

$$\phi'(x) = \left\{ \begin{array}{ll} \{n+1\} & \mathrm{if} x = a\... ...;\mathrm{and}\; x \leq a\\ \phi(x) & \mathrm{otherwise}\\ \end{array}\right.$$

Since $a$ was chosen to be a minimal element, the only relations involving $a$ are of the form $a \leq x$, and for these, by definition, $\phi'(a) = \{n+1\} \subseteq \phi(x) \cup \{n+1\} = \phi'(x)$. For all $x$ such that $a \not\leq x$, we have $\phi'(a) = \{n+1\} \not\subseteq \phi(x)$ by choice of $n$. For relations not involving $a$, the show below that the set containment relations are preserved. Let $\phi(x) \subseteq \phi(y)$. Since $\phi'(y) \supseteq \phi(y)$, the case when $\phi'(x) = \phi(x)$ is trivial. So assume $\phi(x) \subseteq \phi(y)$ and $\phi'(x) = \phi(x) \cup \{n+1\}$. This implies that $a \leq x$, and $x \leq y$, and therefore $a \leq y$. Thus $\phi'(y)$ would be defined as $\phi(y)\cup \{n+1\}$, and hence $\phi'(x) \subseteq \phi'(y)$.Thus the induction step holds.