Metatheory - University of Cambridge

3. Normal forms 28
(cnf2) Every occurrence of negation has minimal scope;
(cnf3) No conjunction occurs within the scope of any disjunction.
Generally, then, a sentence in CNF looks like this
(±A 1 ∨ . . . ∨ ±A i ) ∧ (±A i+1 ∨ . . . ∨ ±A j ) ∧ . . . ∧ (±A m+1 ∨ . . . ∨ ±A n )
where each A k is an atomic sentence.
Since ‘¬’ is its own dual, and ‘∨’ and ‘∧’ are the duals of each other, it
is immediate clear that if a sentence is in DNF, then its dual is in CNF; and
vice versa. Armed with this insight, we can immediately prove another normal
form theorem:
Theorem 3.5. Conjunctive Normal Form Theorem. For any sentence,
there is a tautologically equivalent sentence in conjunctive normal form.
Proof. Let S be any sentence. Applying the DNF Theorem to S, there is a
DNF sentence, D, such that S ⊨ D. So, by Theorem 2.7, S ⊨ D. Whence,
by Lemma 2.5, S ⊨ D. But, since D is in DNF, D is in CNF.
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⊨
⊨
This slick proof is a further illustration of the power of duality. However, it
might suggest that the DNF Theorem enjoys some kind of ‘precedence’ over
the CNF Theorem. That would be misleading. We can easily prove the CNF
Theorem directly, using either of the two proof techniques that we used to
prove the DNF Theorem (whereupon the DNF Theorem could be proved as a
consequence of the CNF Theorem and duality). I shall sketch the main ideas
of the direct proofs of the CNF Theorem; I leave it as an exercise to make these
ideas more precise.
Proof sketch of CNF Theorem via truth tables. Given a TFL sentence, S, we
begin by writing down the complete truth table for S.
If S is true on every line of the truth table, then S ⊨ (A 1 ∨ ¬A 1 ).
If S is false on at least one line of the truth table then, for every line on
the truth table where S is false, write down a disjunction (±A 1 ∨ . . . ∨ ±A n )
which is false on (and only on) that line. Let C be the conjunction of all of
these disjuncts; by construction, C is in CNF and S ⊨ C .
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Proof sketch of CNF Theorem via substitution. Using Lemmas 3.2–3.3, we can
obtain a sentence satisfying (cnf1) and (cnf2). To turn this into a sentence
in CNF, we apply a modified version of Lemma 3.4, obtained by invoking the
duals (what else!) of the Distribution Laws invoked there; i.e.:
((A ∧ B) ∨ C )
(A ∨ (B ∧ C ))
⊨
⊨
⊨
⊨
⊨ ((A ∨ C ) ∧ (B ∨ C ))
⊨ ((A ∨ B) ∧ (A ∨ C ))
⊨
This allows us to pull conjunctions outside the scope of any disjunction.
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Of course, after obtaining a sentence in CNF – however we chose to do it – we
can clean-up the resulting sentence using rules similar to those of §3.4. I leave
it as an exercise, to determine what the appropriate rules should be.