```
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 in `s` 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 possibility `i` in `s`, there will be just as many extensions `i[x/d]` in the new state `s'` as there are entities `d` which in the possible world of `i` 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](/hints/assignment_7_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` in `u`, we collect `dpm`s that are the result of passing through their `bool`, but extending their input `(r, h)` by allocating a new peg for entity `d`, for each `d` in our whole domain of entities, and binding the variable `x` to the index of that peg. A later step can then filter out all the `dpm`s where the entity `d` 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]], then `u` 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 what `fun c -> List.assoc c r` would have been in [week7](
/reader_monad_for_variable_binding).)
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](/reader_monad_for_variable_binding), \\[[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 returns `true` 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](/hints/assignment_7_hint_6). But try to get as far as you can on your own.