-* 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, h) -> let obj = List.nth h (r 'x') in Q obj`. When `...Q obj` evaluates to `true`, that `(r, h)` 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:
-
- bind_set u (fun a -> if test a then unit_set a else empty_set)
-
- Hence, updating `s` with \[[Qx]] should be:
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- bind_set s (fun (r, h) -> if (let obj = List.nth h (r 'x') in Q obj) then unit_set (r, h) else empty_set)
-
- We can call the `(fun (r, h) -> ...)` part \[[Qx]] and then updating `s` with \[[Qx]] will be:
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- bind_set s \[[Qx]]
-
- or as it's written using Haskell's infix notation for bind:
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- 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:
-
- <pre><code>s >>= \[[∃x]] >>= \[[Px]]
- </code></pre>
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- 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 object has the property P will be eliminated.
-
- We can defer that to a later step, where we do `... >>= \[[Px]]`.
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- > 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 objects `d` which in the possible world of `i` have the property P.
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- Deferring the "property P" part, this says:
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- <pre><code>s updated with \[[∃x]] ≡
- s >>= (fun (r, h) -> List.map (fun d -> newpeg_and_bind 'x' d) domain)
- </code></pre>
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- That is, for each pair `(r, h)` in `s`, we collect the result of extending `(r, h)` by allocating a new peg for object `d`, for each `d` in our whole domain of objects (here designated `domain`), and binding the variable `x` to the index of that peg.
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- A later step can then filter out all the possibilities in which the object `d` we did that with doesn't have property P.
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- So if we just call the function `(fun (r, h) -> ...)` above \[[∃x]], then `s` updated with \[[∃x]] updated with \[[Px]] is just:
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- <pre><code>s >>= \[[∃x]] >>= \[[Px]]
- </code></pre>
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- or, being explicit about which "bind" operation we're representing here with `>>=`, that is:
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- <pre><code>bind_set (bind_set s \[[∃x]]) \[[Px]]
- </code></pre>
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-* In def 3.1 on p. 14, GS&V define `s` updated with \[[not φ]] as:
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- > { i &elem; s | i does not subsist in s[φ] }
-
- where `i` *subsists* in <code>s[φ]</code> if there are any `i'` that *extend* `i` in <code>s[φ]</code>.
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- Here's how we can represent that:
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- <pre><code>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
- </code></pre>
-