* Friday, 2011-11-04 at 19:37 -0400 - Martin Bays <mbays@sdf.org>: > So perhaps it would be helpful to think about it like this: > > Given e.g. a binary predicate P(x,y), which let's say is to start with > defined only when x is a foo and y is quux, > (i) has us define what P(X,Y) means where X is a bunch of foos and Y is > a bunch of quuxs (here a bunch of foos corresponds to a set of foos); > meanwhile, (iii) (or something like it) has us define what P(x/~, y/~) > means, where x/~ is an imaginary foo - i.e. one of the new things we get > when we consider a new, coarser notion of equality of foos - and y/~ is > an imaginary quux. > > So we need to consider the properties of these new beasties X and x/~. > One possible, arguably natural, scheme for this in the case of x/~ leads > to the quantifier-permuting ambiguities discussed at the top of this > post. One quick hopefully-clarificatory remark: the "default" semantics of bunches and imaginaries would be conjunctive and disjunctive respectively. i.e. given a predicate P(x,y), unless there were something special going on with P (like with bevri), we would expect that P(X,Y) holds iff P(x,y) does whenever x is in the bunch X and y is in the bunch Y (i.e. P is "distributive"), and P(x/E,y/F) will hold iff P(x',y') does for *some* x' and y' such that x'Ex and y'Fy, i.e. x'/E = x/E and y'/F = y/F, i.e. x' is equivalent to x and y' is equivalent to y where the notions of equivalence are specified by the imaginaries. So e.g. if Beret is the imaginary we get when we consider *all* berets to be equivalent, then if every french person wears a beret we would indeed expect to conclude that every french person "wears" Beret. Just as not all predicates are distributive (in all places), we might expect this disjunctive default to be overwritten sometimes for imaginaries, as in "pure-kind" predication. Martin
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