* How shall we handle \[[∃x]]. As we said, GS&V really tell us how to interpret \[[∃xPx]], but what they say about this breaks naturally into two pieces, such that we can represent the update of `s` with \[[∃xPx]] as:
s >>= \[[∃x]] >>= \[[Px]]
	
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]]`. > 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'` and there are entities `d` which in the possible world of `i` have the property P. Deferring the "property P" part, this says:
s updated with \[[∃x]] ≡
		s >>= (fun (r, h) -> List.map (fun d -> newpeg_and_bind 'x' d) domain)
	
That is, for each pair `(r, h)` in `s`, we collect the result of extending `(r, h)` by allocating a new peg for entity `d`, for each `d` in our whole domain of entities (here designated `domain`), and binding the variable `x` to the index of that peg. A later step can then filter out all the possibilities in which the entity `d` we did that with doesn't have property P. So if we just call the function `(fun (r, h) -> ...)` above \[[∃x]], then `s` updated with \[[∃x]] updated with \[[Px]] is just:
s >>= \[[∃x]] >>= \[[Px]]
	
or, being explicit about which "bind" operation we're representing here with `>>=`, that is:
bind_set (bind_set s \[[∃x]]) \[[Px]]
	
* In def 3.1 on p. 14, GS&V define `s` updated with \[[not φ]] as: > { i &elem; s | i does not subsist in s[φ] } where `i` *subsists* in s[φ] if there are any `i'` that *extend* `i` in s[φ]. Here's how we can represent that:
bind_set s (fun (r, h) ->
			let u = unit_set (r, h)
			in let descendents = u >>= \[[φ]]
			in if descendents = empty_set then u else empty_set