Decreasing quantifiers and entailment in event semantics

Question

Consider the sentences

(1) All dogs bark.

(2) All dogs bark loudly.

In event semantics the logical form of (1) is (∃e,x) every(x) & dog(x) & bark(e,x). As for (2), one would add the conjunct loud(e). The latter formula implies the former, as one would expect. Now consider

(3) Few dogs bark.

(4) Few dogs bark loudly.

Again, the logical form of (4) implies that of (3) (since A & B ⊃ A). But in this case, (3) entails (4) because "few" is a decreasing quantifier. Is there a simple solution to this "paradox"?

Update: a note on the notation (to avoid confusion): In Davidsonian event semantics, sentences are represented as (existential closures of) conjunctions of literals. In dog(x), x can be a dog or a set of dogs (or some others individual that can be lexically described as dog). In this so-called conjunctivist approach, every(x) implies that x isn't a specific dog, but - as some put it - a typical/generic individual defined by the eventuality dog(e,x). There's one axiom in the theory that say that if something holds of a generic element of a set, there's an eventuality with the same predicate for every member of that set. Formally, genericel(x,s) & y∈s & P(e,x) ⊃ (∃e′)P(e′,y). This technique is widely used in commonsense reasoning, but as one can see, there are problems with some quantifiers.

Answer 1

There's no easy way to fix this.

It's a fairly well-known problem and it tends to happen when dealing with "flat" context-free representations -- like event semantics. Jerry Hobbs (of discourse analysis fame) provides one possible solution here. I'm not sure how I feel about it, though. He seems to reference his work on abduction (where there's a knowledge base, KB -- formally, we can represent this as a set of sentences -- that utterances can back-chain to).

In the above case, KB would contain things like all(..., [e]) ⊃ few(..., [e]) but not few(..., [e]) ⊃ all(..., [e]) that enapsulate the quirkiness of of monotone-increasing quantifiers (that retain truth value) and monotone-decreasing quantifiers (that do not necessarily retain truth value). There are also other considerations (such as conservativity; read more here) which need to be taken into account if trying to escape context-free semantics.

There may be better approaches to solving your "paradox", but I am mostly familiar with Hobbs' work.

Answer 2

If

few x are P <-> ~(most x are P)

then, since

most dogs bark loudly -> most dogs bark

we get

~(most dogs bark) -> ~(most dogs bark loudly)

which is equivalent to

few dogs bark -> few dogs bark loudly

and not the other way around .