How to translate these sentences in discourse representation theory How would you translate these sentences in DRT?


Paul has a donkey. He does not like it. 
If a student does not have a cat, she does not have a dog either.


Would that be something like:

Ax[(donkey(x)/\have(paul,x)) -> ~ love(paul,x)]

AxAy[(cat(x) /\ ~have(student,x) &lt;-> (dog(y)/\~have(student,y)
Since you posted here at math.stackexchange.net, I assume you wanted to talk about this as a logic problem. You will need to ask at a https://linguistics.stacktheory.com
if you want to know about modeling this with DRT (Discourse Representation Theory).

You said:

Ax[(donkey(x)/\have(paul,x)) -> ~ love(paul,x)]

which by the way, is totally confusing like and love, so I assume you made a typo.

I'll assume you meant:
Ax[(donkey(x)/\have(paul,x)) -> ~ like(paul,x)]

I read this as meaning the same as:

(for-all x such-that donkey(x) is true
          and where have(paul,x) is true
then 
    (not like(paul,x))) is true.

In English, I would say that this means
every donkey X that satisfies have(paul,X)
must also satisfy (not like(paul,X)) also.

when I read the statement:
   •Paul has a donkey. He does not like it. 

I see several interpretations: 
The first sentence doesn't seem to be subject to 
more than one interpretation.
  Paul has a donkey. 
  Let's give the name Donkey-01 to that donkey.
  so Paul has Donkey-01.
  ie have(Paul,Donkey-01) is true.

So far so good. 

Just for grins and giggles, let's say that Paul has another donkey.
According to the Open World Assumption, we can assume that other facts
are true beyond those we start with.  

Let's give the name Donkey-99 to the second donkey.
Since it is "another donkey" from Donkey-01, we know
that Donkey-01 is different from Donkey-99.

so we know have(Paul,Donkey-01) is true and have(Paul,Donkey-99) is true.
and different(Donkey-01,Donkey-99) is also true.

But the statement "He does not like it" could have multiple interpretations.

Conventionally it would be understood as Paul does not like a particular donkey he has. 
We called the first donkey Donkey-01. So we can assume that Donkey-01 is the donkey the first sentence speaks about. So we can assume that Donkey-01 is also the donkey that Paul
does not like.  So (not like(Paul,Donkey-01)) is true.

If the second sentence is not talking about Donkey-99, we have no information whether Paul likes Donkey-99 or if (not likes(Paul,Donkey-99)

In passing, Unconventionally, the sentence may be saying the Donkey does not like Paul. We have no idea about what gender Paul may be (could be an "it") and we have no idea if the donkey is a "he". So strictly, we can't guarantee the conventional meaning is meant.

We could use a mu expression to make some of this more formal, but I think just using names is sufficient in this case.  

Now technically, "he" and "it" are variables (some folk call them indexicals or even Deictic constructions, but let's not get distracted)
so formally:  the value of the variable he is Paul, and the value of the 
variable "it" is "Donkey-01".  and so we know using the variables, the sentence (not like(he,it)) is true.  But maybe bringing in variables is too fast.

The real issue is that you said for-all X when you meant there-exists.
The original sentence did not say Paul does not like all donkeys. Conventionally understood, the sentence said Paul does not like a particular donkey that he has.
It didn't say anything about other donkeys Paul has, so you can't use a for-all.

As to your second sentence, you need a variable for the student. You used a class when you said ~have(student,x)  which doesn't mean what you think it means.

That third variable (let's call it s) must be the same student which you will use with the sentence (cat(x) &amp;&amp; have(s,x) iff dog(y) &amp;&amp; have(s,y))
and you have said that for-all cats and for-all dogs and should be saying for-all students as well. As well as the obvious that no cat can be a dog and no dog can be a cat. (We assume that is true, but it isn't an axiom in this system)  You also know that the students you care about are female as "she" is mentioned, so you might want to put that in your equation too.

A way to turn this on its head is to use two variables each of which mean all the things that a student may have (let's call the variables h1 and h2), and say that for-all students(s) such-that for-all h1 such-that for-all h2 such-that have(s,h1) &amp;&amp; have(s,h2) &amp;&amp; different(h1,h2) &amp;&amp; not(cat(h1) &amp;&amp; dog(h2)).
Note you don't need to say not(cat(h2) &amp;&amp; dog(h1)) since you didn't restrict h1 and h2 so much that the statement couldn't be used with a different binding for h1 and h2.
Since you posted here at math.stackexchange.net, I assume you wanted to talk about this as a logic problem. You will need to ask at a https://linguistics.stacktheory.com
if you want to know about modeling this with DRT (Discourse Representation Theory).

You said:

Ax[(donkey(x)/\have(paul,x)) -> ~ love(paul,x)]

which by the way, is totally confusing like and love, so I assume you made a typo.

I'll assume you meant:
Ax[(donkey(x)/\have(paul,x)) -> ~ like(paul,x)]

I read this as meaning the same as:

(for-all x such-that donkey(x) is true
          and where have(paul,x) is true
then 
    (not like(paul,x))) is true.

In English, I would say that this means
every donkey X that satisfies have(paul,X)
must also satisfy (not like(paul,X)) also.

when I read the statement:
   •Paul has a donkey. He does not like it. 

I see several interpretations: 
The first sentence doesn't seem to be subject to 
more than one interpretation.
  Paul has a donkey. 
  Let's give the name Donkey-01 to that donkey.
  so Paul has Donkey-01.
  ie have(Paul,Donkey-01) is true.

So far so good. 

Just for grins and giggles, let's say that Paul has another donkey.
According to the Open World Assumption, we can assume that other facts
are true beyond those we start with.  

Let's give the name Donkey-99 to the second donkey.
Since it is "another donkey" from Donkey-01, we know
that Donkey-01 is different from Donkey-99.

so we know have(Paul,Donkey-01) is true and have(Paul,Donkey-99) is true.
and different(Donkey-01,Donkey-99) is also true.

But the statement "He does not like it" could have multiple interpretations.

Conventionally it would be understood as Paul does not like a particular donkey he has. 
We called the first donkey Donkey-01. So we can assume that Donkey-01 is the donkey the first sentence speaks about. So we can assume that Donkey-01 is also the donkey that Paul
does not like.  So (not like(Paul,Donkey-01)) is true.

If the second sentence is not talking about Donkey-99, we have no information whether Paul likes Donkey-99 or if (not likes(Paul,Donkey-99)

In passing, Unconventionally, the sentence may be saying the Donkey does not like Paul. We have no idea about what gender Paul may be (could be an "it") and we have no idea if the donkey is a "he". So strictly, we can't guarantee the conventional meaning is meant.

We could use a mu expression to make some of this more formal, but I think just using names is sufficient in this case.  

Now technically, "he" and "it" are variables (some folk call them indexicals or even Deictic constructions, but let's not get distracted)
so formally:  the value of the variable he is Paul, and the value of the 
variable "it" is "Donkey-01".  and so we know using the variables, the sentence (not like(he,it)) is true.  But maybe bringing in variables is too fast.

The real issue is that you said for-all X when you meant there-exists.
The original sentence did not say Paul does not like all donkeys. Conventionally understood, the sentence said Paul does not like a particular donkey that he has.
It didn't say anything about other donkeys Paul has, so you can't use a for-all.

As to your second sentence, you need a variable for the student. You used a class when you said ~have(student,x)  which doesn't mean what you think it means.

That third variable (let's call it s) must be the same student which you will use with the sentence (cat(x) &amp;&amp; have(s,x) iff dog(y) &amp;&amp; have(s,y))
and you have said that for-all cats and for-all dogs and should be saying for-all students as well. As well as the obvious that no cat can be a dog and no dog can be a cat. (We assume that is true, but it isn't an axiom in this system)  You also know that the students you care about are female as "she" is mentioned, so you might want to put that in your equation too.

A way to turn this on its head is to use two variables each of which mean all the things that a student may have (let's call the variables h1 and h2), and say that for-all students(s) such-that for-all h1 such-that for-all h2 such-that have(s,h1) &amp;&amp; have(s,h2) &amp;&amp; different(h1,h2) &amp;&amp; not(cat(h1) &amp;&amp; dog(h2)).
Note you don't need to say not(cat(h2) &amp;&amp; dog(h1)) since you didn't restrict h1 and h2 so much that the statement couldn't be used with a different binding for h1 and h2.