(* the reason for returning a triple with () in first position will emerge *)
in ((), r',g')
-* At the top of p. 13 (this is in between defs 2.8 and 2.9), GS&V give two examples, one for \[[∃xPx]] and the other for \[[Qx]]. In fact it will be easiest for us to break \[[∃xPx]] into two pieces, \[[∃x]] and \[[Px]]. Let's consider expressions like \[[Px]] (or \[[Qx]]) first.
+* At the top of p. 13 (this is in between defs 2.8 and 2.9), GS&V give two examples, one for \[[∃xPx]] and the other for \[[Qx]]. In fact it will be easiest for us to break \[[∃xPx]] into two pieces, \[[∃x]] and \[[Px]]. Let's consider expressions like \[[Px]] first.
+
+ They say that the effect of updating an information state `s` with the meaning of "Qx" should be to eliminate possibilities in which the object associated with the peg associated with the variable `x` does not have the property Q. In other words, if we let `Q` be a function from objects to `bool`s, `s` updated with \[[Qx]] should be `s` filtered by the function `fun (r, g) -> let obj = List.nth g (r 'x') in Q obj`.
+
+ Recall that [we said before](/hints/assignment_7_hint_2) that `List.filter (test : 'a -> bool) (u : 'a set) : 'a set` is the same as:
+
+ bind_set u (fun a -> if test a then unit_set a else empty_set)
+
+ Hence, updating `s` with \[[Qx]] should be:
+
+ bind_set s (fun (r, g) -> if (let obj = List.nth g (r 'x') in Q obj) then unit_set (r, g) else empty_set)
+
+ We can call the `(fun (r, g) -> ...)` part \[[Qx]] and then updating `s` with \[[Qx]] will be:
+
+ bind_set s [[Qx]]
+
+ or as it's written using Haskell's infix notation for bind:
+
+ s >>= [[Qx]]
+
+* Now 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]]
+
+