(* 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 formula `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`.
+ 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: