Computability and Logic

138 MODELS
There are actually several different sentences that could be used for I n . A comparatively
short one is the following:
∀x 1 ∀x 2 ···∀x n−1 ∃x n (x n ≠ x 1 & x n ≠ x 2 & ... & x n ≠ x n−1 ).
Thus, for instance, I 3 may be written ∀x∀y∃z(z ≠ x & z ≠ y). For this to be true in an
interpretation M, it must be the case that for every p in the domain, if we added a constant
c denoting p, then ∀y∃z(z ≠ c & z ≠ y) would be true. For that to be true, it must
be the case that for every q in the domain, if we added a constant d denoting q, then
∃z(z ≠ c & z ≠ d) would be true. For that to be true, it must be the case that for some r
in the domain, if we added a constant e denoting r, then e ≠ c & e ≠ d would be true. For
that, e ≠ c and e ≠ d would both have to be true, and for that, e = c and e = d would both
have to be untrue. For that, the denotation r of e must be different from the denotations p
and q of c and d. So for every p and q in the domain, there is an r in the domain different
from both of them. Starting from any m 1 in the domain, and applying this last conclusion
with p = q = m 1 , there must be an r, which we call m 2 , different from m 1 . Applying the
conclusion again with p = m 1 and q = m 2 , there must be an r, which we call m 3 , different
from m 1 and m 2 . So there are at least three distinct individuals m 1 , m 2 , m 3 in the
domain.
The set Ɣ of all sentences I n has only infinite models, since the number of elements
in any model must be ≥ n for each finite n. On the other hand, any finite subset Ɣ 0
of Ɣ has a finite model, and indeed a model of size n, where n is the largest number
for which I n is in Ɣ. Can we find an example of a finite set of sentences that has
only infinite models? If so, then we can in fact find a single sentence that has only
infinite models, namely, the conjunction of all the sentences in the finite set. In fact,
examples of single sentences that have only infinite models are known.
12.2 Example (A sentence with only infinite models). Let R be a two-place predicate.
Then the following sentence A has a denumerable model but no finite models:
∀x∃y Rxy& ∀x∀y ∼ (Rxy & Ryx)&∀x∀y∀z((Rxy & Ryz) → Rxz).
A has a denumerable model in which the domain is the natural numbers and the interpretation
of the predicate is the usual strict less-than order relation on natural numbers. For every
number there is one it is less than; no two numbers are less than each other; and if one
number is less than a second and the second less than a third, then the first is less than the
third. So all three conjuncts of A are true in this interpretation.
Now suppose there were a finite model M of A. List the elements of |M| as m 0 ,
m 1 ,...,m k−1 , where k is the number of elements in |M|. Let n 0 = m 0 . By the first conjunct
of A (that is, by the fact that this conjunct is true in the interpretation) there must be some
n in |M| such that R M (n 0 , n). Let n 1 be the first element on the list for which this is
the case. So we have R M (n 0 , n 1 ). But by the second conjunct of A we do not have both
R M (n 0 , n 1 ) and R M (n 1 , n 0 ), and so we do not have R M (n 1 , n 0 ). It follows that n 1 ≠ n 0 .
By the first conjunct of A again there must be some n in |M| such that R M (n 1 , n). Let n 2
be the first element on the list for which this is the case, so we have R M (n 1 , n 2 ). By the