# On donkey sentences: why is this formalization incorrect?

Part of the difficulty surrounding donkey sentences, to my understanding, is about how hard they are to translate to FOL in a matter that is consistent with other translations to FOL in english.

Take "every man who owns a donkey beats it".

The knee-jerk translation would look something like this:

``````∀x[(MAN(x) ∧ ∃y[DONKEY(y) ∧ OWNS(y,x)]) -> BEATS(y,x)]
``````

This is problematic because y is free in the consequent. But now say that we extend the scope of the existential quantifier so it reads as follows:

∀x(MAN(x) -> ∃y[ ([DONKEY(y) ∧ OWNS(y,x)] ∧ BEATS(y,x)) ∧ DONKEY(y) ])

``````∀x(MAN(x) ∧ ∃y[ ([DONKEY(y) ∧ OWNS(y,x)] ∧ BEATS(y,x)) ∧ DONKEY(y) ])
``````

What I did here is extend the scope of the existential to encapsulate the BEATS predicate. Next, I included a conjunction that included another instance of the predicate DONKEY to make the formula more rigorous (because it would evaluate as true if we interpreted y as a pig/non-donkey object). Finally, I moved the conditional after the MAN predicate to "tidy up" the universal and existential quantifiers.

From my intuition, the sentence "every man who owns a donkey beats it" doesn't suffer from ambiguity unless you interpret "beats it" as masturbating. So what's wrong with my translation into formal logic? Am I missing the point of donkey sentences?

Edit: I suspect that this stems from the basic fact that it's discouraged to say ∀x(Px&Qx) compared to ∀x(Px->Qx), but I'm not completely sure why asides from the fact that it's "too strong a claim".

• Why is y being free in the consequent a problem? Also, would there be a problem with ∃y[ DONKEY(y) -> ∀x[(MAN(x) ∧ OWNS(y,x)]) -> BEATS(y,x)] ]? (These are genuine questions, not suggested solutions. My semantics knowledge is very basic) Jan 3 '15 at 2:51
• To my understanding, y isn't bound to the existential quantifier, so it could be interpreted as a different object. Please correct me if I'm wrong (since my semantics knowledge is also limited). Jan 3 '15 at 3:03
• That depends on what kind of quantifiers you're using. How do you distinguish between presuppositions and assertions, for instance? Sep 17 '15 at 16:06

Your formula means that every man owns a donkey that he beats, which is not what the original sentence means.

• Duh, how did that slip my mind! I understand that this is discouraged, but you can make a much stronger claim and simply change the conditional to a conjunction as follows: ∀x(MAN(x) ∧ ∃y[ ([DONKEY(y) ∧ OWNS(y,x)] ∧ BEATS(y,x)) ∧ DONKEY(y) ]). What's the issue here? Jan 3 '15 at 3:06
• That means "Everything whatsoever is a man who owns a donkey he beats", again not what the original English sentence means and really not something that makes much sense to assert anyway. Clearly, not everything is a man. Jan 3 '15 at 3:17
• Maybe in my world! Thanks for clearing all that up. Jan 3 '15 at 3:34

The English sentence "every man who owns a donkey beats it" can be interpreted as "every man beats every donkey that he owns", which makes no implication that anything exists. It is true even if there are no men, no donkeys, or no donkey slavery. This means a rigorous wording of this sentence uses only universal quantifiers ("for all" or "for each"):

For each man x and each donkey y owned by x, x beats y.

Move the universal quantifiers ("for each") to the front:

For all x, for all y, if x is a man, and y is a donkey, and x owns y, then x beats y.

(Poor Lampwick.)

This can be transformed using the definition of implication (if A then B means B or not A):

For all x, for all y, either x beats y, or it is not the case that x is a man, and y is a donkey, and x owns y.

Flattening the negated conjunction with De Morgan's law:

For all x, for all y, either x beats y, or x is not a man, or y is not a donkey, or x does not own y.

This also works with the slang interpretation of "beats it", with "x beats y" replaced with "x stimulates himself".

Another user left a comment pointing out a third interpretation of the sentence, which I find less likely, that leaves at least some donkeys owned by richer men unabused: "every man who owns a donkey beats at least one donkey that he owns." This uses a third free variable: one for the man, a second to identify him as a donkey owner, and a third to identify a particular donkey as a beaten one. This third takes an existential quantifier.

For each man x, for each donkey y owned by x, there exists a donkey z that x owns and beats.

For all x, for all y, if x is a man, y is a donkey, and x owns y, then there exists z where z is a donkey, x owns z, and x beats z.

Perhaps this is why natural language processing is beating researchers' asses, so to speak: one has to capture the stereotypes associated with actors in a sentence in order to resolve this ambiguity.

• Your last formulation might be read: "Every man beats every donkey that he owns." Sep 17 '15 at 20:00
• @GregLee Thanks. I've added a formula for the 3-variable interpretation. Sep 17 '15 at 20:54

The problematic example is difficult to interpret because it assumes with no obvious grounds that no one owns more than one donkey. Taking into account that some may, actually, own several donkeys, it should have been:

"Every man who owns a donkey beats one of the donkeys that he owns."

which clarifies that the owned donkey that is beaten is not necessarily the same donkey as any specific donkey that is owned by the person who does the beating. The unbound variable y for the poor beaten donkey turns up through sloppiness. A distinct variable is needed.