* 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`.
+ 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`. When `...Q obj` evaluates to `true`, that `(r, g)` pair is retained, else it is discarded.
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:
We can call the `(fun (r, g) -> ...)` part \[[Qx]] and then updating `s` with \[[Qx]] will be:
- bind_set s [[Qx]]
+ bind_set s \[[Qx]]
or as it's written using Haskell's infix notation for bind:
- s >>= [[Qx]]
+ 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]]
+ s >>= \[[∃x]] >>= \[[Px]]