@MISC{NORMFORM.MIZ,
  AUTHOR = {Trybulec, Andrzej},
  TITLE = {Algebra of Normal Forms},
  DAY = {05},
  MONTH = {October},
  YEAR = {1990},
  ADDRESS1 = {Warsaw University\\Bia{\l}ystok},
  SUMMARY = {We mean by a normal form a finite set of ordered pairs of subsets of a
fixed set that fulfils two conditions: elements of it consist of disjoint
sets and element of it are incomparable w.r.t. inclusion. The underlying
set corresponds to a set of propositional variables but it is arbitrary.
The correspondents to a normal form of a formula, e.g. a disjunctive normal
form is as follows. The normal form is the set of disjuncts and a disjunct
is an ordered pair consisting of the sets of propositional variables that
occur in the disjunct non-negated and negated. The requirement that
the element of a normal form consists of disjoint sets means that
contradictory disjuncts have been removed and the second condition means
that the absorption law has been used to shorten the normal form.
We construct a lattice $\langle {\Bbb N},\sqcup,\sqcap \rangle$,
where $ a \sqcup b = \mu (a \cup b)$ and
 $a \sqcap b = \mu c$, $c$ being set of all pairs
 $\langle X_1 \cup Y_1, X_2 \cup Y_2 \rangle$, $\langle X_1, X_2 \rangle
 \in a$ and $\langle Y_1,Y_2 \rangle \in b$, which consist of disjoiny
sets. $\mu a$ denotes here
the set of all minimal, w.r.t. inclusion, elements of $a$.
We prove that the lattice of normal forms over a set defined in this way
is distributive and that $\emptyset$ is the minimal element of it.}}

