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[Again, the "box" is just the universal quantifier]
Kamp, Hans, van Genabith, and Reyle (forthcoming) offer a fairly thorough account of the problems with donkey sentences. What they consider first is the standard method for translating natural languages sentences into first-order predicate logic. The existential quantifier is defined as follows:
(1) $xF Û Ø“x ØF
Consider a simple sentence:
(2) A delegate arrived.
Translated into FOL we get:
(3) $x(delegate(x) Ù arrived(x))
(3) is logically equivalent to:
(4) Ø“x Ø(delegate(x) Ù arrived(x))
When (4) is translated into English we get:
(5) It is not the case that every delegate failed to arrive.
Now consider a sentence with anaphora:
(6) A delegatei arrived. Shei registered.
Kamp, Hans, van Genabith, and Reyle introduce the superscripts and subscripts in order to keep track which noun matches up with which pronoun.
If we apply what we did in steps (1)-(5) to (6) we get the following sentence:
(7) It is not the case that every delegatei failed to arrive. Shei registered.
Intuitively, there is something wrong with (7) as it does not capture what is expressed by (6).
References
Kamp, H., J. van Genabith and U. Reyle, forthcoming. “Discourse Representation Theory”, in Gabbay D. and F. Guenthner, Handbook of Philosophical Logic (second edition), Springer.
[I don't know why, but the "all" quantifier is coming out as a box. I have seen that before, but usually the "some" quantifer also comes out as a box which it isn't in this case. Just read the box as an "all" quantifier]
Well, it has been a while since I posted. I think I am still recovering from Spring 2008 semester. Well, whatever the reason (or excuse) is, I hope to start posting again. I have been trying to work out the problem known as “Donkey sentences.” I realize when discussing the topic with some of my colleagues, they have trouble trying to figure out what the problem. Therefore, I have been trying to more clearly state the problem than the readings I have come across.
The problem of what have come to be called “Donkey sentences” has to do with a certain kind of anaphora (statements about other statements) has received considerable attention since it was reintroduced in Peter Geach’s Reference and Generality. The problem is often introduced as follows, consider the following sentences:
(1) If Pedro owns a donkey then he beats it.
(2) Every farmer who owns a donkey beats it.
What has taken to be the “natural” strong reading of those sentences is: (1) is the case only if Pedro beats all the donkeys that he owns and (2) is the case only if every farmer beats the donkey(s) she/he owns. The natural readings of (1) and (2) has created considerable problems when trying to translate them into First-Order-Logic (FOL) as no FOL translation can capture these natural readings. There have been two important responses to this problem, Discourse Representation Theory (DRT) proposed by Hans Kamp in 1981 and Independence Friendly Logic (IF Logic) proposed by Jaakko Hintikka in 1985. Both solutions have had little discussion between the two of them and so both have developed rather independently of each other.[1]
Since both offer different solutions to the same problem it should be determined which (if either) side offers a better solution to the problem. After reviewing both sides I will argue that IF Logic has a better solution to the problem because it is more in line with FOL and so is a simpler and less ad hoc solution. My strategy for approaching this problem will be as follows; first I will sketch out a more precise account of the problem. Second, I will summarize DRT’s solution and IF Logic’s solution. Third, I will give my reasons for claiming that IF is a simpler and less ad hoc solution.
II. The Problem with Donkey Sentences
There have been attempts to translate sentences (1) and (2) into FOL that can capture the truth conditions already mentioned. With respect to (1), three translations have been offered:
(1a) $x[donkey(x) Ù (owns(pedro, x) ® beats(pedro, x))]
(1b) $x[(donkey(x) Ù owns(pedro, x))] ® beats(pedro, x)
(1c) “x[(donkey(x) Ù owns(pedro, x)) ® beats(pedro, x)]
Similar translations have been attempted with respect to (2),
(2a) “x[(farmer (x) Ù $y(donkey(y) Ù owns(x, y))) ® beats(x, y)]
(2b) “x$y[(farmer (x) Ù donkey(y) Ù (owns(x, y) ® beats(x, y))]
(2c) “x“y[(farmer (x) Ù donkey(y) Ù owns(x, y)) ® beats(x, y)]
The problem with (1b) is fairly obvious as it does not express a well-formed formula (wff) because it leaves a free occurrence of the bound variable x. Why the remaining propositions fail needs a little more explanation.
The reason (1a) fails is the proposition comes out true whenever there is a donkey that Pedro doesn’t happen to own. Demonstrating that (1a) comes out true in such an instance is fairly simple given the definition of the conditional and conjunction. The conditional comes out false if and only if (iff) the antecedent is true and the consequence false and true otherwise. The definition of conjunction says that it will come out true iff both conjuncts are true and false otherwise. Suppose that x is a donkey, but it is a donkey that is not owned by Pedro. Such a state of affairs means the conditional in (1a) will come out true as the antecedent (owns(pedro, x)) will be false therefore the conjunct (owns(pedro, x) ® beats(pedro, x)) comes out true. Since x is a donkey then the other conjunct (donkey(x)) also comes out true. Therefore both conjuncts are true; therefore the proposition is true. This fails to capture the natural reading of (1) as (1) is supposed to only come out true when Pedro beats all the donkeys he owns.
The reason (1c) fails is the proposition comes out true when Pedro owns something other than a donkey. Suppose that Pedro owns a pig and ignore whether he beats it or not. Such a state of affairs means that one of the conjuncts in the antecedent is false (donkey(x)). Therefore, the conjunction (donkey(x) Ù owns(pedro, x)) is also false. Therefore, the conditional is true as it has a false antecedent. If it is the case that Pedro can own a pig and this ownership will make (1c) come out true then clearly this fails to capture the natural reading of (1).
[1] There is one exception to this, in “No Scope for Scope?” Hintikka does give some reason to accept IF over DRT, but it is very brief and difficult to understand. I will attempt to clarify Hintikka’s argument later in the paper.
