Bulletin of the Section of Logic
Volume 39:3/4 (2010), pp. 199–204
Jacob Vosmaer
A NEW VERSION OF AN OLD MODAL
INCOMPLETENESS THEOREM
Abstract
Thomason [5] showed that a certain modal logic L ⊂ S4 is incomplete with
respect to Kripke semantics. Later Gerson [3] showed that L is also incomplete
with respect to neighborhood semantics. In this paper we show that L is in fact
incomplete with respect to any class of complete Boolean algebras with operators,
i.e. that it is completely incomplete.
1.
Introduction
In 1974, two modal incompleteness theorems were published in the same
issue of the same journal. Fine [2] presented a logic above S4 and Thomason
[5] presented one between T and S4, and both showed that their logics were
incomplete with respect to Kripke semantics. In 1975, a paper by Gerson [3]
followed in which he showed that both logics were both also incomplete with
respect to neighborhood semantics. Then, in 2003 Litak [4] showed that
Fine’s logic is in fact as he calls it completely incomplete, i.e. it is incomplete
with respect to any class of complete Boolean algebras with operators (or
BAOs for short). It is known that Kripke frames correspond to the class of
complete, atomic and completely additive BAOs, and that neighborhood
frames for normal logics such as the ones we are considering correspond to
the class of complete, atomic BAOs (see Thomason [6] and Došen [1]). In
the present paper, we show what one might call a complement to Litak’s
result, viz. that Thomason’s logic is also completely incomplete.
200
2.
2.1.
Jacob Vosmaer
An incompleteness theorem
Algebraic preliminaries
When considering an arbitrary complete BAO A below, we will always assume there is some Kripke frame hW, Ri such that A = hA, ∧, −, 0, ♦i is
a subalgebra of h℘(W ), ∩, c , ∅, mR i, where c is set-theoretic complementation with respect to W and for U ⊆ W , mR (U ) := {w ∈ W | ∃v ∈
U (wRv)}; the Jónsson-Tarski theorem tells us that any BAO is such a
subalgebra up to isomorphism. We will make use of a fewSobservations
W
about suprema in a complete BAO A. Let X, Y ⊆ A. By X and X
we denote the supremum of X in P(W ) and in A, respectively.S Because
W A
is not a regular subalgebra of P(W ), it may be the case thatS X 6= W X.
Firstly, note however that it will always be the case that X ≤ X.
Secondly,
[
X≤
[
Y implies
_
X≤
_
Y,
(1)
S
S
W
W
as for all a ∈ X, aW≤ X ≤ Y ≤ Y , so Y is an upper bound for X
in A. Now
W
W since X is the least upper bound of X in A, it follows that
X ≤ Y . Thirdly,
[
X∩
[
Y = ∅ implies
_
X∧
_
Y = 0,
(2)
S
S
W
W
W
S
for if X ∩ Y = ∅ but X ∧ SY > 0Wthen W X ∩ Y > 0. If this were
not the case,
W then we would get Y ⊆ Y \ X ∈ A, contradicting the
fact that Y is the W
least upper bound of Y in A. So, there
must
S
W be some
b ∈ Y suchWthat b ∩ X > 0, and now we know that X * X \ b, for
otherwise X would not be the least upper bound of X in A. It follows
that there must be some
X such that a∧b > 0; however this contradicts
S a ∈S
our assumption that X ∩ Y = ∅. It follows that (2) is true. Finally,
_
a∈X
♦a ≤ ♦
_
a.
(3)
a∈X
W
Take any
it follows that
Wa ∈ X, then a ≤ X. Since ♦ is order-preserving,
W
♦a ≤ ♦ X. Since a was arbitrary, we see that ♦ X is an upper bound
for {♦a | a ∈ X}. It follows that (3) holds.
A New Version of an Old Modal Incompleteness Theorem
2.2.
201
A case of complete incompleteness
Consider the axioms
Ai := (qi → r)
(i = 1, 2),
Bi := (r → ♦qi )
(i = 1, 2),
C1 := ¬(q1 ∧ q2 ),
A := r ∧ p ∧ ¬2 p ∧ A1 ∧ A2 ∧ B1 ∧ B2 ∧ C1
→ ♦(r ∧ (r → q1 ∨ q2 )),
B := (p → q) → (p → q),
C := p → p,
D := (p ∧ ♦2 q) → (♦q ∨ ♦2 (q ∧ ♦p)),
E := (p ∧ ¬2 p) → ♦(2 p ∧ ¬3 p),
F := p → 2 p.
Let L be the logic containing all propositional tautologies, A, B, C, D and
E and closed under modus ponens, substitution and necessitation. This is
the same logic as found in [5]; note that B, C and F are perhaps better
known as the K-axiom, the T -axiom and the 4-axiom respectively.1 It is
therefore not hard to see that T ⊆ L ⊆ S4. We will see below that the
latter inclusion is strict, because S4 3 F ∈
/ L.
Lemma 1. Let A be a complete BAO. If A |= L, then A |= F .
Proof: Let A be a complete BAO on which B, C, D and E are valid2 , but
F is not. We will show that A 6|= A, proving the statement of the lemma.
The fact that A 6|= F must be witnessed by some a ∈ A such that
a 2 a. Since by C, 2 a ≤ a, it follows that 2 a < a. For n ≥ 1
we define
bn := n a \ n+1 a,
1 Incidentally, axiom B plays no role in the proof of our result, nor does the fact that
the modalities of BAOs are additive. We only use the facts that (and hence also ♦)
is monotone and that 1 = 1. By assuming B we are following Thomason and ensuring
that T ⊆ L.
2 Using the algebraizability of modal logic, we view modal formulas as equations,
e.g. the formula A corresponds to the equation A = 1.
202
Jacob Vosmaer
where c \ d := c ∧ −d. By the above, we already know that b1 > 0. To
inductively show that all bn > 0, suppose that bn > 0, but bn+1 = 0. Then
replace3 p with n−1 a in E, so we get
bn = n−1 a ∧ −(2 n−1 a) ≤ ♦(2 n−1 a ∧ −(3 n−1 a)) = ♦bn+1 =
♦0 = 0,
which is a contradiction, so it must be that bn+1 > 0. This completes our
induction. Note that
for all 1 ≤ i < j,
bi ∧ bj = 0,
(4)
since bj ≤ j a ≤ i+1 a ≤ −bi . Next, we will show by induction on n that
for all n ≥ 2 and for all 1 ≤ i < n,
bi ≤ ♦bn .
(5)
The base case n = 2 follows immediately from E. We will show that if (5)
holds for n, it must also hold for n + 1. We only consider i < n (for if i = n,
we can immediately apply E). By our induction hypothesis, bi ≤ ♦bn and
by E, bn ≤ ♦bn+1 , so we have bi ≤ ♦2 bn+1 , i.e. bi = bi ∧ ♦2 bn+1 . Reverting
to definitions, we find that ♦bi = ♦(i a\i+1 a). As i a\i+1 a ≤ −i+1 a,
we get that ♦(i a \ i+1 a) ≤ ♦ − i+1 a = −i+2 a, so i+2 a ∧ ♦bi = 0.
Since also bn+1 ≤ n+1 a ≤ i+2 a (as i < n), it follows that bn+1 ∧ ♦bi = 0,
so replacing p with bi and q with bn+1 in D, we find that
bi = bi ∧ ♦2 bn+1 ≤ ♦bn+1 ∨ ♦2 (bn+1 ∧ ♦bi ) = ♦bn+1 ∨ ♦2 0 = ♦bn+1 .
It follows that (5) holds for n + 1, so by induction (5) is true for all n ≥ 2.
Now we define the following elements of A:
_
_
p := a, qi :=
b3n+i (i = 1, 2, 3), r :=
bn .
n≥0
n≥1
(Note that this is where we use the assumption that A is complete.)
We will
S
use
these
elements
to
show
that
A
is
not
valid.
First
of
all,
as
b
n≥0 3n+i ≤
S
b
,
it
follows
by
(1)
that
q
≤
r
for
i
=
1,
2,
3,
so
q
→
r
=
1, whence
n
i
i
n≥1
A1 = A2 = 1 = 1. Secondly, by (5), for any n ≥ 1 there must exist a
3 0 a
:= a.
A New Version of an Old Modal Incompleteness Theorem
203
S
S
k ≥ 2 such that bn ≤ ♦b3k+i , whence n≥1 bn ≤ n≥0 ♦b3n+i . By (1), this
means that
_
_
_
for all 1 ≤ i ≤ 3, r =
bn ≤
♦b3n+i ≤ ♦
b3n+i = ♦qi ,
(6)
n≥1
n≥0
n≥0
where the latter inequality follows from (3). Therefore, r S
→ ♦qi = 1,
so
B
=
B
=
1
=
1.
Finally,
using
(4),
we
see
that
1
2
n≥0 b3n+i ∩
S
b
=
0
for
all
1
≤
i
<
j
≤
3.
It
follows
by
(2)
that
3n+j
n≥0
for all 1 ≤ i < j ≤ 3,
qi ∧ qj = 0
(7)
whence C1 = − 0 = 1. Combining all this, we find that
r ∧ p ∧ −2 p ∧ A1 ∧ A2 ∧ B1 ∧ B2 ∧ C1 = r ∧ (a \ 2 a) = b1 > 0,
i.e. the left hand side of A is non-zero. However, we have r = q1 ∨ q2 ∨ q3 .
Since the qi are disjoint by (7), this means that r ∧ −q1 ∧ −q2 = q3 . By
(6), r ≤ ♦q3 , so
0 = r ∧ −♦q3 = r ∧ − q3 = r ∧ − (r ∧ −q1 ∧ −q2 ) = r ∧ (r → q1 ∨ q2 ).
It follows that ♦(r ∧ (r → q1 ∨ q2 )) = 0, i.e. the right hand side of A is
zero. We conclude that A 6|= A.
For C some class of BAOs, we define ∆ |=C Γ if for every A ∈ C, A |= ∆
only if A |= Γ.
Corollary 2. Let C be any class of complete BAOs. Then {A, B, C, D, E}
|=C F .
Lemma 3. F ∈
/ L.
Proof: See the proof of the Theorem in [5]. Thomason proves our lemma
by showing that the veiled recession frame, which is a general frame equivalent to an incomplete BAO, validates L while ¬F can be satisfied on
it.
The lemmas give us the following:
Theorem 4. L is completely incomplete.
204
3.
Jacob Vosmaer
Acknowledgements
This paper is the result of Eric Pacuit asking me to write a paper about
neighborhood semantics for modal logic for a class of his in January of
2006 at the Institute for Logic, Language and Computation. He was very
helpful although the result strayed somewhat from the original assignment.
I would also like to thank fellow students Gaëlle Fontaine and Christian
Kissig for discussions. Finally, I would like to express my gratitude to the
anonymous referee for their comments and suggestions.
References
[1] K. Došen, Duality between modal algebras and neighborhood frames, Studia
Logica 48 (1989), pp. 219–234.
[2] Kit Fine, An incomplete logic containing S4, Theoria 40 (1974), pp. 23–29.
[3] Martin Gerson, The inadequacy of the neighborhood semantics for modal
logic, The Journal of Symbolic Logic 40 (1975), pp. 141–148.
[4] Tadeusz Litak, Modal incompleteness revisited, Studia Logica 76 (2004),
pp. 329–342.
[5] S. K. Thomason, An incompleteness theorem in modal logic, Theoria 40
(1974), pp. 30–34.
[6] S. K. Thomason, Categories of frames for modal logic, The Journal of
Symbolic Logic 40 (1975), pp. 439–442.
Universiteit van Amsterdam
Institute for Logic, Language and Computation
Postnus 94242, 1090 GE Amsterdam, The Netherlands
e-mail: [email protected]