From 30535a559f5df2bcca773bc9b292cbc8e8896174 Mon Sep 17 00:00:00 2001 From: Jim Pryor Date: Thu, 18 Nov 2010 17:51:11 -0500 Subject: [PATCH 1/1] assignment7 tweaks Signed-off-by: Jim Pryor --- hints/assignment_7_hint_3.mdwn | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/hints/assignment_7_hint_3.mdwn b/hints/assignment_7_hint_3.mdwn index 61f48a71..7cc2e74a 100644 --- a/hints/assignment_7_hint_3.mdwn +++ b/hints/assignment_7_hint_3.mdwn @@ -29,7 +29,7 @@ More specifically, \[[expression]] will be a set of `'a discourse_possibility` m * 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: -- 2.11.0