We generalize the following results of infinitary first order logic:
A class of models is axiomatizable by universal / existential / positive / negative formulas iff it is closed under submodels / extensions / homomorphic images / surjective homomorphic preimages.
Assume we are given a class of ‘models’ and a class
of formulas, forming a (complete) Boolean algebra (so that operations ‘and’/’or’ are defined for arbitrary many formulas), and assume we are given a complete Boolean relation
between them (for ‘validity’), i.e. a relation satisfying
and
.
Let .
We write and
.
A class of models is axiomatizable by
-formulas if
for some
, equivalently, if
.
Let be a reflexive binary relation on the class of models (typically stating the existence of a morphism of certain type), and write
, similarly define
.
Definition. A binary relation is modelwise expressible by
-formulas, if for every model
there is a formula
which is valid on
, and whenever
, it implies
.
Examples.
Let the formulas be infinitary first order formulas for a similarity type , and the models be the corresponding relational / algebraic structures.
- Let
mean that
is a homomorphic image of
, i.e. isomorphic to a quotient of
, and choose
to be the class of positive formulas. Set
whereis the conjunction of all atomic predicates that hold in
, the element
replaced by the corresponding variable
.
- Let
mean that
is isomorphic to a substructure of
, i.e. there is a relationally full embedding
, and choose
to be the class of existential formulas. Set
whereruns over atomic formulas which are not valid on
under the substitution
.
Theorem. Suppose is modelwise expressible by
-formulas. Then
a) If , then
is axiomatizable by
-formulas.
b) If , then
is axiomatizable by disjunctions of
-formulas.
Proof.
a) Let and
. As
, we have
yielding a
with
which shows by hypothesis that
.
b) Let and
. We want to show
.
Define and
. Then
, so
, meaning that there is an
for which
.
Now use the hypothesis for model : there is a
, so that
. But it can’t be in
because
, so
, yielding
thus
.
This theorem proves one direction in all of the above mentioned preservation theorems.
The converse direction, that the specific types of formulas are preserved under the specific operator on the class of models is a straightforward verification in each case.
However, we can prove a common generalization for them, using the more specific logic described in this post.
Let be an arbitrary profunctor with an initial object
, objects of
regarded as ‘situations’ and objects of
as ‘models’, and introduce the relation ‘diagonal fill-in‘ (or ‘left lifting property‘) between arrows of
and arrows of
as follows:
that is, iff every commutative square with on the top and
on the bottom has a diagonal fill-in
making the whole diagram commutative:
For a subclass of morphisms




For example, in the first order case, we have surjective morphisms of
, and
injective morphisms of
.
Moreover, whenever
, and
whenever
and
consists only of equations.
Universal formulas correspond to those trees in which Eve
doesn’t select new variables (so that the part doesn’t change in arrows on odd levels), and positive formulas correspond to trees in which
Adam
doesn’t select new relations or equations to hold (so that the part is empty in the root situation and doesn’t change in arrows on even levels).
Theorem. Let be an arbitrary class of morphisms in
, and let
be a tree in which every arrow
Eve
can possibly choose is in . Then
implies
.
Dually, if is of the above form, then
implies
.
It turns out that examples 1. and 2. above have a common generalization in this setting.
Theorem. Let be a functor, surjective on hom-sets, suppose
has pushouts and assume that a class
of morphisms is defined by a set
of arrows by the diagonal-fill-in relation:
.
Then the relation between objects stating
is modelwise expressible by trees in which the arrows that
Adam
might choose are all in the profunctor
.
Proof. Since is opfunctorial, every
has a coreflection arrow
, which is now also a pre-reflection: each heteromorphism
factors through (not necessarily uniquely)
, because of the surjectivity condition.
For all and arrow
, fix a pushout square
and collect all such
‘s in a class
. Note that from each
there starts out only set-many arrows of
, and also note that
holds because
is a pushout of
.
For a given , consider the following tree:
where every arrow



Adam
) and the next layer of the tree consists of all right inverses 

Proving is straightforward after letting
Eve
choose the coreflection arrow .
Now, assume for an arbitrary object
. We want to show the existence of an arrow
.
Let’s call the interpretation
Eve
chooses on . By hypothesis,
is also a pre-reflection arrow, yielding an arrow
such that
. We show
.
Let and assume a commutative square
is given. Since
is a coreflection arrow, there is a (unique)
such that
. Then use the chosen pushout of
and
to form
and let
Adam
choose branch 


Eve
‘s answer branch with interpretation forced to be 



This is going to work as a diagonal-fill-in in the originally given square
