Consider
1. Sam and Dave are meeting together.
2. Al, Bill, and Carl are meeting together.
3. Some people are meeting together.
Obviously, neither (1) nor (2) can be decomposed into a conjunction of singular predications. Thus (2) cannot be analyzed as 'Al is meeting together & Bill is meeting together & Carl is meeting together.' So it is natural to try to analyze (1) and (2) using relational predicates. (1) becomes
1R. Meeting(Sam, Dave) In symbols: Msd
But if 'meeting' is a dyadic (two-place) predicate, then we should expect (2) to give way to
2R. Mab & Mbc & Mac.
Unfortunately, (2R) is true in circumstances in which (2) is false. Suppose there are three separate meetings. Then (2R) is true and (2) false. To get around this difficulty, we can introduce a triadic relation M* which yields as analysans of (2):
2R*. M*abc.
But then we need a tetradic relation should Diana come to the meeting. And so on, with the result that 'meeting together' picks out a family of relations of different polyadicities. But what's wrong with that? Well, note that (1) and (2) each entail (3) by Existential Generalization in the presence of the auxiliary premise 'Al, Bill, Carl, Dave, and Sam are people.'
But then we are going to have difficulty explaining the validity of the two instances of Existential Generalization. For the one instance features a dyadic meeting relation and the other a triadic. If two different relations are involved, then what is the logical form of (3) -- Some people are meeting together -- which is the common conclusion of both instances of Existential Generalization? If 'meeting together picks out a family of relations of different 'adicities, then (3) has no one definite logical form.
Does this convince you that the multiple relations approach is unworkable?
REFERENCE: Thomas McKay, Plural Predication (Oxford 2006), pp. 19-21.
Looks like we need plural quantification, with new variables picking out plural objects: the x's. The variable will have to range over sets of objects with cardinality > 1, i.e., in our toy world over the members of M:
M = {(sam, dave), (al, bill, cal)}
So we have: there are x's such that Mx's.
Posted by: Boram Lee | Saturday, July 31, 2010 at 02:07 PM
Sorry I missed this post. This answers the question in my comment on the previous post. It seems as though the multiple relations approach is unworkable.
Posted by: William | Saturday, July 31, 2010 at 02:22 PM
What follows is quite obvious, I will nonetheless write it. Let (Mn) denote the family of relations in question indexed by natural numbers (greater than two if one is afraid of meeting oneself). 1., 2. and 3. can then be reformulated in our model as
1*. M2( S , D ) obtains.
2*. M3( A , B , C ) obtains.
3*. There exist a natural number n and persons P1, ... , Pn s.t. Mn( P1, ... , Pn ) obtains.
3*. is clear and the inferences from 1*. to 3*. and 2*. to 3*. are valid. Is 3*. unintuitive? Hardly. When we say that some people met, we mean that some particular number of people met, even if we do not know the number. A meeting in which no particular number of people took part is an incoherent notion. How can one blame the model for not making sense of 3. irrespectively of this number? Is there an obvious problem with 3* that I am missing?
Posted by: Jan | Saturday, July 31, 2010 at 03:15 PM
>>3*. There exist a natural number n and persons P1, ... , Pn s.t. Mn( P1, ... , Pn ) obtains.
>>Is there an obvious problem with 3* that I am missing?
Not at all, but I thought that the 'relational' model was one where we could replace Mn(a,b,c ...) with a set of dyadic relations M1(a,b) & M2(a,c) and so on. The point is that there appear to be some predicates which are irreducibly of the adicity which they have. (Perhaps I misunderstood your point).
Posted by: William | Saturday, July 31, 2010 at 11:14 PM
Well, the first part of the original post proposes a simple Binary Relation Model, while the second part a Family of Relations Model (FoRM). Both are found wanting. I disagree with dr. Vallicella's assessment of FoRM. Here are the relevant passages:
"And so on, with the result that 'meeting together' picks out a family of relations of different polyadicities. But what's wrong with that? Well, note that (1) and (2) each entail (3) by Existential Generalization in the presence of the auxiliary premise 'Al, Bill, Carl, Dave, and Sam are people.' But then we are going to have difficulty explaining the validity of the two instances of Existential Generalization."
No difficulty that I can see.
"For the one instance features a dyadic meeting relation and the other a triadic. If two different relations are involved, then what is the logical form of (3) -- Some people are meeting together -- which is the common conclusion of both instances of Existential Generalization? If 'meeting together picks out a family of relations of different 'adicities, then (3) has no one definite logical form."
3*. is quite definite.
"Does this convince you that the multiple relations approach is unworkable?"
No.
Posted by: Jan | Sunday, August 01, 2010 at 10:28 AM
Jan,
Your 3* looks a perfectly respectable way of expressing 'some people are meeting together' but I suspect it is second-order since it appears to quantify over the relations M2, M3, M4... as well as the persons P1, P2,... McKay's objective, as I understand it, is to enhance the expressive power of first-order languages whilst retaining their nice logical properties.
Posted by: David Brightly | Sunday, August 01, 2010 at 11:27 AM
Jan,
Excellent comment which presents a nice challenge.
1. Sam and Dave are meeting together.
2. Al, Bill, and Carl are meeting together.
3. Some people are meeting together.
1*. M2( S , D ) obtains.
2*. M3( A , B , C ) obtains.
3*. There exists a natural number n and persons P1, ... , Pn s.t. Mn( P1, ... , Pn ) obtains.
I have no objection to your symbolization of (1) and (2). The question is whether (3*) gives us the logical form of (3) given that (3) is the first-order existential generalization of (1) and of (2) respectively.
As David B points out, (3*) is second-order since it quantifies over relations using the relation-variable 'Mn.' 'M2' and 'M3' by contrast are relation-constants.
So your (3*) is not the first-order existential generalization of (1*) or of (2*).
Second-order E.G. may be problematic. From 'Bill is happy' one can validly infer 'Someone is happy.' That's first-order E. G. But from 'Bill is happy' can one validly infer 'There is a property P such that B instantiates P'? That's 2nd order E.G. I think William would scream bloody murder.
Can one infer validly 'There is a person x and a property P such that x instantiates P'?
As I understand McKay, his logic remains first-order.
Posted by: Bill Vallicella | Sunday, August 01, 2010 at 01:35 PM
>>But from 'Bill is happy' can one validly infer 'There is a property P such that B instantiates P'?< For 'B' substitute 'Bill.'
If 2nd order E.G. is kosher, then (3*) does follow from (1*) and (2*) respectively. But there is nothing 2nd order about (3). The predicate in (3) is a relation-constant.
Posted by: Bill Vallicella | Sunday, August 01, 2010 at 01:40 PM
Bill,
See what you think of this simple solution. Mab is a symmetric, non-reflexive, non-transitive relation representing the utterance "A and B are in a meeting together." We can extend Mab to M*abc and define M*abc dyadically, but we must recognize a change in Mab to M*ab, where M* is now transitive (such that if a is now in a meeting with b, he is also in a meeting with anyone else now meeting with b). Thus M*abc = M*ab and M*bc and M*ac. In English M*ab is a little complicated. It says something like a and b are now in a meeting, possibly with other people, and in any case there is only one meeting now involving a and b and possibly others.
Posted by: Philoponus | Monday, August 02, 2010 at 08:41 AM
Your notation seems confused. So I'm not understanding you. 'M' picks out a dyadic meeting relation. 'M*' picks out a triadic meeting relation. Accordingly, "Thus M*abc = M*ab and M*bc and M*ac" doesn't make sense.
In any case, didn't I refute your suggestion above by pointing out that (2) is not captured by (2R)?
I must be missing something.
Posted by: Bill Vallicella | Monday, August 02, 2010 at 12:26 PM