Tools and Techniques in Modal Logic Marcus Kracht II ...

1.6. Properties of Logics 27
theorem which allows to prove interpolation for logics based on a finite matrix. Say
that ⊢ has a conjunction if there is a term p ∧ q such that the following are derivable
rules: 〈{p, q}, p ∧ q〉 and both 〈{p ∧ q}, p〉 and 〈{p ∧ q}, q〉. In addition, if ⊢ = ⊢M
for some logical matrix we say that ⊢ has all constants if for each t ∈ T there exists
a nullary term–function t such that for all valuations v v(t) = t. (Note that since
var(t) = ∅ the value of t does not depend at all on v.) This rather complicated
definition allows that we do not need to have a constant for each truth–value; it
is enough if they are definable from the others. For example in classical logic we
may have only ⊤ = 1 as a primitive and then 0 = ¬⊤. Notice also the following.
Say that an algebra is functionally complete if every function A n → A is a term–
function of A; and say that A is polynomially complete if every function A n → A
is a polynomial function. Then any functionally complete algebra is polynomially
complete; the converse need not hold. However, if A has all constants, then it is
functionally complete iff it is polynomially complete.
Theorem 1.6.4. Suppose that M is a finite logical matrix. Suppose that ⊢M has
a conjunction ∧ and all constants; then ⊢M has interpolation.
Proof. Suppose that ϕ(�p, �q) ⊢ ψ(�q,�r), where �r = 〈ri : i < n〉. Clearly, var(ψ) �
var(ϕ), iff n > 0. We show that if n � 0 there exists a
ψ 1 = ψ 1 (�q, r0, . . . rn−2)
such that ϕ ⊢ ψ1 ⊢ ψ. The claim is then proved by induction on n. We put
ψ 1 �
(�q, r0, . . . , rn−2) := 〈ψ(�q, r0, . . . , rn−2, t) : t ∈ T〉
Now observe that ϕ ⊢ ψ implies
ϕ(�p, �q) ⊢ ψ(�q, r0, . . . , rn−2, t)
for every t. Now we apply the rule for conjunction for each t ∈ T, and obtain
ϕ(�p, �q) ⊢
�
〈ψ(�q, r0, . . . , rn−2, t) : t ∈ T〉 (= ψ 1 ) .
Furthermore, ψ 1 ⊢ ψ, that is,
�
〈ψ(�q, r0, . . . , rn−2, t)〉 ⊢ ψ(�q, r0, . . . , rn−1) .
For if v(ψ 1 ) is true then for any extension v + with rn−1 ∈ dom(v + ) we have v + (ψ) ∈
D. �
The following is an instructive example showing that we cannot have interpolation
without constants. Consider the logic of the 2–valued matrix in the language
C ′ = {∧, ¬}, with 1 the distinguished element. This logic fails to have interpolation.
For it holds that p ∧ ¬p ⊢ q, but there is no interpolant if p � q. This is surprising
because the algebra 〈2, −, ∩〉 is polynomially complete. Hence, polynomial completeness
is not enough, we must have functional completeness. In other words, we
must have all constants. Let us also add that the theorem fails for intersections of
logics with all constants and conjunction.