Metatheory - University of Cambridge

3. Normal forms 25
Step 1: Remove all the double-negations in the sentence (working from left to
right in the sentence).
Step 2: Starting from the left, look for the first instance where we can apply one
of the De Morgan laws, and apply it, pushing negations inside brackets.
Step 3: Repeat steps 1–2, until no further change can occur.
It might help to see this algorithm in action, on the example from before:
(¬(¬A ∨ (¬B ∧ C)) ∨ ¬((D ∧ E) ∨ (¬D ∧ ¬E)))
((¬¬A ∧ ¬(¬B ∧ C)) ∨ ¬((D ∧ E) ∨ (¬D ∧ ¬E)))
((A ∧ ¬(¬B ∧ C)) ∨ ¬((D ∧ E) ∨ (¬D ∧ ¬E)))
((A ∧ (¬¬B ∨ ¬C)) ∨ ¬((D ∧ E) ∨ (¬D ∧ ¬E)))
((A ∧ (B ∨ ¬C)) ∨ ¬((D ∧ E) ∨ (¬D ∧ ¬E)))
((A ∧ (B ∨ ¬C)) ∨ (¬(D ∧ E) ∧ ¬(¬D ∧ ¬E)))
((A ∧ (B ∨ ¬C)) ∨ ((¬D ∨ ¬E) ∧ ¬(¬D ∧ ¬E)))
((A ∧ (B ∨ ¬C)) ∨ ((¬D ∨ ¬E) ∧ (¬¬D ∨ ¬¬E)))
((A ∧ (B ∨ ¬C)) ∨ ((¬D ∨ ¬E) ∧ (D ∨ ¬¬E)))
((A ∧ (B ∨ ¬C)) ∨ ((¬D ∨ ¬E) ∧ (D ∨ E)))
We are really very close to DNF! All that remains is to meet condition (dnf3),
i.e. to ensure that no disjunction occurs within the scope of any conjunction.
So our final task is to prove:
Lemma 3.4. For any sentence satisfying (dnf1) and (dnf2), there is a tautologically
equivalent sentence in DNF.
Proof. We shall almost do a (strong) induction on length. However, the only
measure of length that we care about is the number of occurrences of disjunctions
in a sentence. More precisely, then, the disjunctive-length of
a sentence is the number of occurrences of ‘∨’ in the sentence, and we shall
proceed by (strong) induction on disjunctive-length. 2
Let A be any sentence satisfying (dnf1) and (dnf2). For induction, suppose
that, for every sentence B which satisfies (dnf1) and (dnf2) and is
disjunctively-shorter than A (i.e. that contains fewer disjunctions than A),
there is a DNF sentence B ∗ which is exactly as disjunctively-long as B and
such that B and B ∗ . There are now four cases to consider:
Case 1: A is atomic. In this case, A itself is in DNF and is (trivially)
exactly as disjunctively-long as A.
Case 2: A is ¬B. In this case, B must be atomic, hence ¬B is in DNF
and is (trivially) exactly as disjunctively-long as A.
Case 3: A is (B ∨ C ). Then A ⊨ (B ∗ ∨ C ∗ ), which is in DNF and, by
assumption, is exactly as disjunctively-long as A.
Case 4: A is (B ∧ C ). There are three subcases to consider, depending
upon the form of B ∗ and C ∗ :
⊨
2 I leave it as an exercise, to check that the required Principle of Strong Induction on
Disjunctive-Length is legitimate; compare the justification of the Principle of Strong Induction
on Length in §1.4.