# A. Proof of Theorem 4.1

Theorem A.1   Let be any finite partial order. Let be the lattice of subsets of with the set inclusion ordering. Then there exists a mapping , such that and is a finite subset of for all .

Proof : We prove the theorem by induction on .
Base Case : Let = 1. Then the claim trivially holds.
Induction Hypothesis : Let the claim hold for all such that .
Induction Step : .
Let be a partial order such that . Since is finite, has a minimal element, say . Consider the partial order . Clearly this is a partial order and . Hence by induction hypothesis, there exists , such that and is a finite subset of for all . Let . Define as follows.

Since was chosen to be a minimal element, the only relations involving are of the form , and for these, by definition, . For all such that , we have by choice of . For relations not involving , the show below that the set containment relations are preserved. Let . Since , the case when is trivial. So assume and . This implies that , and , and therefore . Thus would be defined as , and hence .Thus the induction step holds.