Metatheory - University of Cambridge

2. Substitution 18
We shall use the same kind of proof-technique in what follows, to great effect.
First, we need a little more notation, connecting duality with negation.
Given any sentence, A, we say A is the result of replacing every connective
in A with its dual, and A is the result of replacing every atomic sentence in
A with the negation of that atomic sentence. So, when A is the sentence
‘(A ∨ (¬B ∧ C))’, A is ‘(A ∧ (¬B ∨ C))’ and A is ‘(¬A ∨ (¬¬B ∧ ¬C))’.
Theorem 2.6. A
⊨
⊨ ¬A, for any sentence A.
Proof. Let A be any sentence; we suppose, for induction on length, that every
shorter sentence B is such that A ⊨ ¬A. There are two cases to consider:
⊨
Case 1: A is atomic. Then A is just A itself; and ¬A is just ¬¬A; and
we know that A ⊨ ¬¬A.
Case 2: A is ◁(B 1 , . . . , B n ), for some n-place connective ◁. So by hypothesis,
B i ⊨ ¬B i for each B i . Now observe that:
⊨
⊨
◁(B 1 , . . . , B n )
⊨
⊨
⊨
⊨
⊨
⊨ ◁(B 1 , . . . , B n )
by definition
⊨ ◁(¬B 1 , . . . , ¬B n ) by Lemma 2.2
⊨ ¬◁(¬¬B 1 , . . . , ¬¬B n ) by definition of ◁
⊨ ¬◁(B 1 , . . . , B n ) by Lemma 2.2
⊨ ¬◁(B 1 , . . . , B n )
by definition
and hence, given the transitivity of entailment, A
This completes the induction on length.
⊨
⊨ ¬A.
■
And we can now obtain a very elegant result:
Theorem 2.7. A ⊨ B iff B ⊨ A, for any sentences A and B. Hence, in
particular, A ⊨ B iff A ⊨ B.
⊨
⊨
Proof. Left to right. Suppose A ⊨ B. Then A ⊨ B, by Lemma 2.1. So
¬B ⊨ ¬A. So B ⊨ A, by Theorem 2.6 and Lemma 2.2.
Right to left. Suppose B ⊨ A. Then, by the preceding, A ⊨ B. Hence
A ⊨ B, by Lemmas 2.5 and 2.2.
■
This ends my theorem-proving. But to close the chapter, I want to offer some
sketch some reasons for thinking that duality is pretty great.
Illustration 1. You may have already noticed the following equivalences:
(A ∧ (B ∨ C ))
((A ∨ B) ∧ C )
⊨
⊨
⊨ ((A ∧ B) ∨ (A ∧ C ))
⊨ ((A ∧ C ) ∨ (B ∧ C ))
If not, then you can test these using a truth table. But, having done so,
Theorem 2.7 immediately yields two further equivalences for free:
(A ∨ (B ∧ C ))
((A ∧ B) ∨ C )
⊨
⊨
⊨ ((A ∨ B) ∧ (A ∨ C ))
⊨ ((A ∨ C ) ∧ (B ∨ C ))