Metatheory - University of Cambridge

3. Normal forms 26
4a: B ∗ contains a disjunction. In that case, since B ∗ is in DNF, B ∗ is
(B 1 ∨ B 2 ), for some B 1 and B 2 in DNF that are both disjunctivelyshorter
than B ∗ . Next, observe:
(B ∧ C )
⊨
⊨
⊨ ((B 1 ∨ B 2 ) ∧ C )
⊨ ((B 1 ∧ C ) ∨ (B 2 ∧ C ))
Now, (B 1 ∧ C ) is disjunctively-shorter than A; so by hypothesis,
there is some DNF sentence D which is exactly as disjunctivelylong
as (B 1 ∧ C ) and such that D ⊨ (B 1 ∧ C ). Similarly, there
is some DNF sentence E which is exactly as disjunctively-long as
(B 2 ∧ C ) and such that E ⊨ (B 2 ∧ C ). So A ⊨ (D ∨ E), which is
in DNF and exactly as disjunctively-long as A.
4b: C ∗ contains a disjunction. Exactly similar to Subcase 4a; I leave the
details as an exercise.
4c: neither B ∗ nor C ∗ contains a disjunction. Then A ⊨ (B ∗ ∧ C ∗ ),
which is already in DNF and is exactly as long as A itself.
This completes the induction.
⊨
Again, this Lemma implicitly provides an algorithm; this time, for pulling
disjunctions outside the scope of any conjunction, by repeatedly applying the
Distribution Laws:
((A ∨ B) ∧ C )
(A ∧ (B ∨ C ))
⊨
⊨
⊨
⊨
⊨ ((A ∧ C ) ∨ (B ∧ C ))
⊨ ((A ∧ B) ∨ (A ∧ C ))
Applying these to our earlier example, we get, successively:
((A ∧ (B ∨ ¬C)) ∨ ((¬D ∨ ¬E) ∧ (D ∨ E)))
(((A ∧ B) ∨ (A ∧ ¬C)) ∨ ((¬D ∨ ¬E) ∧ (D ∨ E)))
(((A ∧ B) ∨ (A ∧ ¬C)) ∨ (((¬D ∨ ¬E) ∧ D) ∨ ((¬D ∨ ¬E) ∧ E)))
(((A ∧ B) ∨ (A ∧ ¬C)) ∨ (((¬D ∧ D) ∨ (¬E ∧ D)) ∨ ((¬D ∧ E) ∨ (¬E ∧ E))))
And this is in DNF, as may become clearer if we temporarily allow ourselves
to remove brackets, following the notational conventions of fx C§10.3:
(A ∧ B) ∨ (A ∧ ¬C) ∨ (¬D ∧ D) ∨ (¬E ∧ D) ∨ (¬D ∧ E) ∨ (¬E ∧ E)
Together, then, these Lemmas implicitly provide us with an algorithm for computing
an equivalent sentence, in DNF. Moreover, this algorithm – although
lengthy – was much more efficient at generating a DNF sentence than the
truth-table method would have been. (The truth table for our example would
have 32 lines, 22 of which are true.)
It remains to assemble these Lemmas, into a proof of the DNF Theorem:
Proof of DNF Theorem via substitution. Let A be any TFL sentence.
Lemma 3.2 tells us that there is a sentence, B, satisfying (dnf1), such that
A ⊨ B.
Lemma 3.3 tells us that there is a sentence, C , satisfying (dnf1) and (dnf2),
such that B ⊨ C .
Lemma 3.4 tells us that there is a sentence, D, in DNF, such that C ⊨ D.
And, by the transitivity of entailment, A ⊨ D.
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⊨
⊨
⊨
⊨
⊨
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