Computability and Logic

19.1. DISJUNCTIVE AND PRENEX NORMAL FORMS 245
Proof: Given any formula, first replace it by a negation-normal equivalent. Then,
using the distributive laws, that is, the equivalence of (B &(C ∨ D)) to ((B & C) ∨
(B & D)) and of ((B ∨ C)&D) to((B ∨ D)&(C ∨ D)), ‘push conjunction inside’
and ‘pull disjunction outside’ until a disjunctive normal equivalent is obtained.
(It would be a tedious but routine task to rewrite this ‘top down’ description of the
process of finding a disjunctive normal equivalent as a ‘bottom up’ proof the existence
of such an equivalent.)
If in a formula that is in disjunctive normal form each disjunction contains each
A i exactly once, plain or negated, then the compound is said to be in full disjunctive
normal form. (A notion of full conjunctive normal form can be defined exactly analogously.)
In connection with such forms it is often useful to introduce, in addition to
the two-place connectives ∨ and & , and the one-place connective ∼, the zero-place
connectives or constant truth ⊤ and constant falsehood ⊥, counting respectively as
true in every interpretation and false in every interpretation. The disjunction of zero
disjuncts may by convention be understood to be ⊥, and the conjunction of zero conjuncts
to be ⊤ (rather as, in mathematics, the sum of zero summands is understood
to be 0, and the product of zero factors to be 1).
In seeking a full disjunctive normal equivalent of a given disjunctive normal formula,
first note that conjunctions (and analogously, disjunctions) can be reordered
and regrouped at will using the commutative and associative laws, that is, the equivalence
of (B & C)to(C & B), and of (B & C & D), which officially is supposed to be
an abbreviation of (B &(C & D)), with grouping to the right, to ((B & C)&D), with
grouping to the left. Thus for instance (P &(Q & P)) is equivalent to (P &(P & Q))
and to ((P & P)&Q). Using the idempotent law, that is, the equivalence of B & B
to B, this last is equivalent to P & Q. This illustrates how repetitions of the same A i
(or ∼A i ) within a conjunction can be eliminated. To eliminate the occurrence of the
same A i twice, once plain and once negated, we can use the equivalence of B & ∼B to
⊥ and of ⊥ & C to ⊥, and of ⊥∨D to D, so that, for instance, (B & ∼B & C) ∨ D is
equivalent simply to D: contradictory disjuncts can be dropped. These reductions will
convert a given formula to one that, like our earlier example (∼P & Q) ∨ (∼P & R),
is a disjunction of conjunctions in which each basic formula occurs at most once,
plain or negated, in each conjunct.
To ensure that each occurs at least once in each conjunction, we use the equivalence
of B to (B & C) ∨ (B & ∼C). Thus our example is equivalent to
and to
(∼P & Q & R) ∨ (∼P & Q & ∼R) ∨ (∼P & ∼R)
(∼P & Q & R) ∨ (∼P & Q & ∼R) ∨ (∼P&Q & ∼R) ∨ (∼P & ∼Q & ∼R)
and, eliminating repetition, to
(∼P & Q & R) ∨ (∼P & Q & ∼R) ∨ (∼P & ∼Q & ∼R)
which is in full disjunctive normal form. The foregoing informal description can be
converted into a formal proof of the following result.