Presuppositions in Spoken Discourse

Chapter 2
(32)
In the representation in (32) we resolved the pronoun in two steps, by first
introducing a reference marker, and then stipulating what other reference marker in
the representation it is identical with. What we can then later do is replace all
instances of z with y, and then remove z from the list of reference markers. We
end up with a DRS that will have the same truth conditions as (32) above but one
that is easier to read. In the following I will use the more direct method of
immediately resolving pronouns and other anaphoric expressions to the reference
marker they are binding with. Only where it is necessary for clarity will I show the
step-by-step process of identifying an antecedent.
What are the truth-conditions of (32)? This DRS is true if there is a function
between the reference markers in the DRS and a model of interpretation such that
the reference marker x is mapped onto an individual j in the model, and the
reference marker y is mapped onto another individual b in the model, and j is Julia
and b is a bicycle and j and b are related in such as way that j owns b and b is red.
The DRS in (32) therefore has the same truth-conditions as its translation into
predicate logic, which would be the following:
(33) ∃x ∃y ( Julia(x) ∧ bicycle(y) ∧ owns(x, y) ∧ red(y))
The important difference between the predicate logic formula and the DRSstructure
is that the DRS-structure can be updated further with more information.
The predicate logic structure cannot be incrementally changed because the variables
in the structure are sealed off. DRT also has a semantics that allows its
representations to be given a model-theoretic interpretation. DRSs are interpreted
as partial models of the world. A DRS is true in a model if we can find real
individuals in that model that are in the relationship coded by the conditions given
in the DRS being evaluated.
There is a way to write DRSs in a linear format. This makes the
representation a little more difficult to read but it takes up much less space and
contains the same information so I will use the linear format in the following.
Example (34) illustrates this format for the DRS given in (32), but with y
substituted for z.
(34) [ x,y : Julia(x), bicycle(y), owns(x,y), red(y)]
20
x y z
Julia(x)
bicycle(y)
x owns y
z = y
red(z)