Can presentists say enough? Part I: Counting

There are eternalist sentences which seem to make perfect sense, but which it is hard for a presentist to make sense of. For instance, consider this sentence:

1. There were, are and will be a total of n horses,

where n is a finite or infinite cardinality. David Lewis showed that there is a way for presentists to make sense of (1) for finite n, but last time I checked, it wasn't known if this can be done for infinite cardinalities. Of course, if there are haecceities it's easy: (1) is equivalent to saying that the set of haecceities that have been, are or will be exemplified by a horse has cardinality n. But haecceities seem to me to be a cheat, whether for presentists or for modal actualists. Besides, (1) doesn't seem to be about haecceities—it seems to be about horses. So it's an interesting question whether (1) can be made sense of for infinite cardinalities without haecceities.

It turns out it can. For the solution, the presentist needs an "ExistAgo(x,t)" operator, where t is a real number (in some fixed unit system), so that "ExistsAgo(x,7)" means that x existed 7 units ago, "ExistsAgo(x,0)", which we can abbreviate "Exists(x)", means that x presently exists, and "ExistsAgo(x,−3)" means that x will exist in 3 units, and a "t ago:" sentential operator (where t is a real number in the same unit system), with the embedding rules being such that "t ago: u ago: p" is roughly equivalent to "t+u ago: p"[note 1] while "t ago: ExistsAgo(x,u)" is roughly equivalent to "t+u ago: Exists(x)".

For with such operators, we can define counting as follows. Let S be the set of all non-empty subsets of the reals. If K is a kind-term and s is a member of S, there is a unique cardinality n(K,s) such that for any member t of s we have:

2. t ago: n(K,s)=Card { x : Kx & ExistsExactlyAgo(x,s−t) },

where s−t = { u : u=v−t for some v in s }, and ExistsExactlyAgo(x,s) iff for all real t, ExistsAt(x,t) iff t is a member of s. In other words, n(K,s) is the number of Ks that exist exactly at the times indicated by s. Then we say:

3. There have been, are and will be exactly n Ks iff the sum of n(K,s), as s ranges over the members of S, equals n.

So the non-haecceitist presentist can solve the problem of counting items at different times, even up to arbitrary cardinality. Moreover, it seems like here horses really are being counted, not just horse haecceities, though they're being counted in a funny sort of aggregation.

There are, of course, other expressibility challenges for presentists, especially non-haecceitist ones. More on that on a later occasion.