How shall we handle [[∃x]]? As we said, GS&V really tell us how to interpret [[∃xPx]], but for our purposes, what they say about this can be broken naturally into two pieces, such that we represent the update of our starting set
u
with [[∃xPx]] as:u >>= [[∃x]] >>= [[Px]]
(Extra credit: how does the discussion on pp. 25-29 of GS&V bear on the possibility of this simplification?)
What does [[∃x]] need to be here? Here's what they say, on the top of p. 13:
Suppose an information state
s
is updated with the sentence ∃xPx. Possibilities ins
in which no entity has the property P will be eliminated.We can defer that to a later step, where we do
... >>= [[Px]]
. GS&V continue:The referent system of the remaining possibilities will be extended with a new peg, which is associated with
x
. And for each old possibilityi
ins
, there will be just as many extensionsi[x/d]
in the new states'
as there are entitiesd
which in the possible world ofi
have the property P.Deferring the "property P" part, this corresponds to:
u updated with [[∃x]] ≡ let extend one_dpm (d : entity) = dpm_bind one_dpm (new_peg_and_assign 'x' d) in set_bind u (fun one_dpm -> List.map (fun d -> extend one_dpm d) domain)
where
new_peg_and_assign
is the operation we defined in hint 3:let new_peg_and_assign (var_to_bind : char) (d : entity) : bool -> bool dpm = fun truth_value -> fun (r, h) -> (* first we calculate an unused index *) let new_index = List.length h (* next we store d at h[new_index], which is at the very end of h *) (* the following line achieves that in a simple but inefficient way *) in let h' = List.append h [d] (* next we assign 'x' to location new_index *) in let r' = fun var -> if var = var_to_bind then new_index else r var (* we pass through the same truth_value that we started with *) in (truth_value, r', h');;
What's going on in this proposed representation of [[∃x]]? For each
bool dpm
inu
, we collectdpm
s that are the result of passing through theirbool
, but extending their input(r, h)
by allocating a new peg for entityd
, for eachd
in our whole domain of entities, and binding the variablex
to the index of that peg. A later step can then filter out all thedpm
s where the entityd
we did that with doesn't have property P. (Again, consult GS&V pp. 25-9 for extra credit.)If we call the function
(fun one_dom -> List.map ...)
defined above [[∃x]], thenu
updated with [[∃x]] updated with [[Px]] is just:u >>= [[∃x]] >>= [[Px]]
or, being explicit about which "bind" operation we're representing here with
>>=
, that is:set_bind (set_bind u [[∃x]]) [[Px]]
Let's compare this to what [[∃xPx]] would look like on a non-dynamic semantics, for example, where we use a simple Reader monad to implement variable binding. Reminding ourselves, we'd be working in a framework like this. (Here we implement environments or assignments as functions from variables to entities, instead of as lists of pairs of variables and entities. An assignment
r
here is whatfun c -> List.assoc c r
would have been in week7.)type assignment = char -> entity;; type 'a reader = assignment -> 'a;; let reader_unit (value : 'a) : 'a reader = fun r -> value;; let reader_bind (u : 'a reader) (f : 'a -> 'b reader) : 'b reader = fun r -> let a = u r in let u' = f a in u' r;;
Here the type of a sentential clause is:
type clause = bool reader;;
Here are meanings for singular terms and predicates:
let getx : entity reader = fun r -> r 'x';; type lifted_unary = entity reader -> bool reader;; let lift (predicate : entity -> bool) : lifted_unary = fun entity_reader -> fun r -> let obj = entity_reader r in reader_unit (predicate obj)
The meaning of [[Qx]] would then be:
[[Q]] ≡ lift q [[x]] ≡ getx [[Qx]] ≡ [[Q]] [[x]] ≡ fun r -> let obj = getx r in reader_unit (q obj)
Recall also how we defined [[lambda x]], or as we called it before, [[who(x)]]:
let shift (var_to_bind : char) (clause : clause) : lifted_unary = fun entity_reader -> fun r -> let new_value = entity_reader r (* remember here we're implementing assignments as functions rather than as lists of pairs *) in let r' = fun var -> if var = var_to_bind then new_value else r var in clause r'
Now, how would we implement quantifiers in this setting? I'll assume we have a function
exists
of type(entity -> bool) -> bool
. That is, it accepts a predicate as argument and returnstrue
if any element in the domain satisfies that predicate. We could implement the reader-monad version of that like this:fun (lifted_predicate : lifted_unary) -> fun r -> exists (fun (obj : entity) -> lifted_predicate (reader_unit obj) r)
That would be the meaning of [[∃]], which we'd use like this:
[[∃]] ( [[Q]] )
or this:
[[∃]] ( [[lambda x]] [[Qx]] )
If we wanted to compose [[∃]] with [[lambda x]], we'd get:
let shift var_to_bind clause = fun entity_reader r -> let new_value = entity_reader r in let r' = fun var -> if var = var_to_bind then new_value else r var in clause r' in let lifted_exists = fun lifted_predicate -> fun r -> exists (fun obj -> lifted_predicate (reader_unit obj) r) in fun bool_reader -> lifted_exists (shift 'x' bool_reader)
which we can simplify to:
fun bool_reader -> let shifted r new_value = let r' = fun var -> if var = 'x' then new_value else r var in bool_reader r' in fun r -> exists (shifted r)
This gives us a value for [[∃x]], which we use like this:
[[∃x]] ( [[Qx]] )
Contrast the way we use [[∃x]] in GS&V's system. Here we don't have a function that takes [[Qx]] as an argument. Instead we have a operation that gets bound in a discourse chain:
u >>= [[∃x]] >>= [[Qx]]
The crucial difference in GS&V's system is that the distinctive effect of the [[∃x]]---to allocate new pegs in the store and associate variable
x
with the objects stored there---doesn't last only while interpreting some clauses supplied as arguments to [[∃x]]. Instead, it persists through the discourse, possibly affecting the interpretation of claims outside the logical scope of the quantifier. This is how we'll able to interpret claims like:If ∃x (man x and ∃y y is wife of x) then (x kisses y).
See the discussion on pp. 24-5 of GS&V.
Can you figure out how to handle [[not φ]] and the other connectives? If not, here are some more hints. But try to get as far as you can on your own.