Friday, April 29, 2011

formulating a donkey sentence in first-order logic

Here is a standard donkey sentence:

(1) Every farmer that owns a donkey beats it.

It is often said that such sentences cannot be formulated in the language of first-order logic. There is a formulation in first-order logic that gets the correct truth conditions but it differs radically from the surface structure of (1)---most significantly it interprets the indefinite article as a universal quantifier.

(1') ∀x∀y[(farmer(x) & donkey(y) & owns(x,y)) --> beat(x,y)]

There are numerous novel semantic frameworks that attempt to deal with the problem, e.g. various dynamic semantics. But the following seems to be a reasonable formulation of the donkey sentence in first-order logic.

(1'') ∀x[(farmer(x) & ∃y(y=z & donkey(z) & owns(x,z)) --> beat(x,z)]

What's wrong with this?


DilipNinan said...

I don't see how the existential quantifier is affecting the interpretation.

Moreover, if you generally translate indefinites using a free variable which is implicitly universally quantified (which is how I understand what you're doing in (1'')), you'll get the wrong results for "A man walked in the park." Part of the challenge is to give a translation procedure for sentences containing indefinites into first-order logic that is somewhat general. (Btw, Heim discusses various attempts to provide such a procedure in Ch. 1 of her dissertation.)

Brian Rabern said...

hmm, thanks.

I was thinking that "A man walked in the park" would be:

∃y(y=z & man(z) & walked-in-park(z))

But I can't treat the free variable as implicitly existential, while treating the free variable in the donkey sentence as implicitly universal.

I thought I had essentially mimicked the dynamic treatment by using free variables, but I'll have to think more about why there is a difference.