In a posthumous paper, David Lewis shows that one can find a presentist paraphrase of sentences like "There have ever been, are or ever will be *n* *F*s" for any finite *n*. But his method doesn't work for infinite counting.

It turns out that there is a solution that works for finite and infinite counts, using a bit of set theory. For any set *S* of times, say that an object *x* exactly occupies *S* provided that at every time in *S* it was, is or will be the case that *x* exists and at no time outside of *S* it was, is or will be the case that *x* exists. For any non-empty set *S* of times, let *n*_{F}(*S*) be a cardinality such that at every time *t* in *S* it was, is or will be the case that there are exactly *n*_{F}(*S*) objects exactly occupying *S*. This is a presentist-friendly definition. Let *N* be any set of abstracta with cardinality *n*_{F}(*S*) (e.g., if we have the Axiom of Choice, we should have an ordinal of that cardinality) and let *e*_{F}(*S*) be the set of ordered pairs { <*S*,*x*> : *x*∈*N* }. We can think of the members of *e*_{F}(*S*) as the ersatz *F*s exactly occupying *S*. Let *e*_{F} be the union of all the *e*_{F}(*S*) as *S* ranges over all subsets of times. (It's quite possible that I'm using the Axiom of Choice in the above constructions.) Then "There have ever been, are or ever will be *n* *F*s" can be given the truth condition |*e*_{F}|=*n*.

This ersatzist construction suggests a general way in which presentists can talk of ersatz past, present or future objects. For instance, "There were, are or ever will be more *F*s than *G*s" gets the truth condition: |*e*_{G}|≤|*e*_{F}|. "Most *F*s that have ever been, are or will be were, are or will be *G*s" gets the truth condition |*e*_{FG}|>(1/2)|*e*_{F}|, where *F**G* is the conjunction of *F* with *G*. I don't know just how much can be paraphrased in such ways, but I think quite a lot. Consequently, just as I think the B-theory can't be rejected on linguistic grounds, it's going to be hard to reject presentism on linguistic grounds.