# Truth-conditions of predicate-logic formulas for donkey sentences

I'm current learning about compositional semantics, quantifier raising and scope ambiguity in my semantics class and I'm having trouble answering some questions. I've attempted to answer the questions below, but i'm not sure if i'm answering the questions in the correct way.

(1) If Jake owns a donkey, he beats it

(a) The following predicate logic formula does not correctly represent the truth-conditional meaning of (1). Why not? Explain your answer.

(2) ∃x[donkey(x) ∧ own(j, x)] → beat(j, x)

(b) The following predicate logic formula does not correctly represent the truth-conditional meaning of (1) either. Why not? Explain your answer.

(3) ∃x[[donkey(x) ∧ own(j, x)] → beat(j, x)]

(c) Provide a predicate logic formula that most closely represents the truth-conditional meaning of (1). Explain your answer.

a) The formula does not represent the truth conditional meaning because the existential quantifier does not scope over (x) in beat(j,x), meaning that (x) is free. The donkey that Jake owns in own(j,x), is not the same donkey that was beaten in beat(j,x). The truth conditions of the entire sentence would be false, because (x) is not an element of both own(j,x) and beat(j,x).

I'm a bit confused about b) and c), because I thought that b) was the correct representation.

If I could get any help with these questions, I would be very grateful!

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– prash
Commented Jul 25, 2017 at 13:23

a) Correct. Some fine-tuning to your answer:

The donkey that Jake owns in own(j,x), is not the same donkey that was beaten in beat(j,x).

It is not necessarily the same. Depending on which assignment function the formula is evaluated under, the assignment for the second occurence of the variable x can co-incide with the one for the first x. But what is important is, as you figured out, that the first occurence of the variable is in the scope of a quantifier why the second one is not, which means that x and x are not the same variable and thus possibly differ in their denotation.

The truth conditions of the entire sentence would be false, because (x) is not an element of both own(j,x) and beat(j,x).

You need to be careful about syntax and semantics: xis a variable, i.e. some string of symbols, and own(j,x) and beat(j,x) are formulas, i.e. also just some strings of symbols. Syntactically, x does occur in both of the formulas. It is also not meaningful to say that something is an element of (the interpretation of) own(j,x) because own(j,x) is a formula which evaluates to a truth condition, true or false is not something a variable or an individual could be an element of. What you want to say is that the tuples <j,x> and <j,x> are not talking about the same element, i.e., the denotation of x for the first tuple is not necessarily an element of the denotation of the second tuple and vice versa.

b)

∃x[[donkey(x) ∧ own(j, x)] → beat(j, x)]

Carfully look at what the formula says: It says that

there is an something such that if this something is a donkey and owned by Jake, then it is beaten by Jake.

This is not the desired outcome:
By putting an existential quantifier with scope over the entire sentence, the formula forces there to be such a something. This is, however, not implied by original sentence: It just says

If there is a thing which is a donkey and owned by Jake, then...

Having an existential quantifier with scope over the entire sentence thus makes a wrong existential prediction.

What is even worse: Suppose a variable assignment assings x some individual d from the domain, and this d happens not to be a donkey owned by Jake (could be it is not a donkey, could be it doesn't belong to Jake, could be both). We will most likely find such an individual if we assume that our domain consists of more than just Jake's donkeys. As we know, if the antecedent of the implication fails, the implication gets true tirivally, no matter whether or not this d is beaten by Jake. And since we found at least one assignment for which the implication gets true, we're now done with evaluating the existential quantifier and the statement evaluates to true.
This means that as soon as we find one individual which is not a donkey and owned by Jake, the whole formula suddenly becomes true. This is certainly not what we want!
In general, an existentially quantified implication (i.e. ∃x[φ → ψ] is almost always a sign that something is wrong.

c) What you want to say instead is

Look at an arbitrary individual. If this individual is a donkey and owned by Jake, then...

which is equivalent to

For any individual it holds that if it is a donkey and owned by Jake, then it gets beaten by Jake.

The magic lies in the word any: This condition triggers a universally quantified formula. What you thus end up with is

(4) ∀x[[donkey(x) ∧ own(j, x)] → beat(j, x)]

It may seem unintuitive at first sight we we would need a universal quantifier here, but it helps to think about it in terms of the recursive evaluation of truth conditions:
(4) is true if the implication is true for all assignments of the variable x to some individual d in the domain. Now go through the possible assignments step by step: Under the current assignment, is d both a donkey and owned by Jake?
No? Then we don't care - it could be beaten by Jake or it could not; in any case it doesn't matter because we are only interested in whether those individuals that do satisfy the antecedent also satisfy the consequent. Therefore, the implication gets true for those individuals which make the antecedent false.
Yes? Then we found a donkey which is owned by Jake and need to make sure now that it is also beaten by Jake. Thus, for those assignments which deliver an indivdual d that satisfy the antecedent of the implication, the consequent must hold as well. As soon as we find one counterexample (i.e. some d such that d is a donkey and owned by Jake, but not beaten by him), we know that the sentence must be false (because it says that if there is a donkey that is owned by Jake, then it must be beaten by him). If, however, all such d that satisfy the antecedent also satisfy the consequent of the implication, then the sentence is true, no matter how many of them are there.
If these conditions succeed for all assignments, i.e. any individual we look at is either not a donkey of Jake's (in which case we don't care) or it is beaten by Jake (in which case the proposition about Jake's donkeys applies), then the formula becomes true.
Note that this formula now doesn't come with an unwanted existential force either: In case that there just isn't any donkey that is owned by Jake, under all variable assignments the antecedent gets false, the implication true and thus the universally quantified formula is vavuously true - so it doesn't hurt if there just aren't any donkeys that Jake owns, as long as if there is one, it gets beaten, which is what the original sentence states.