Tyrrell_McAllister comments on Notion of Preference in Ambient Control - Less Wrong

14 Post author: Vladimir_Nesov 07 October 2010 09:21PM

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Comment author: Vladimir_Nesov 08 October 2010 08:24:32PM *  0 points [-]

For example, you claim that [A=1] must be derivable in the theory if the agent actually does A.

If the agent actually does 1 (I assume you meant to say). I don't see what you are trying to say again. I agree with the last paragraph (you could recast the argument that way), but don't understand the third paragraph.

Comment author: Tyrrell_McAllister 08 October 2010 08:34:07PM 0 points [-]

If the agent actually does 1 (I assume you meant to say).

Whoops. Right.

I don't see what you are trying to say again. I agree with the last paragraph (you could recast the argument that way), but don't understand the third paragraph.

Okay. Let me try to make my point by building on the last paragraph, then. According to my understanding, you start out knowing that v(A) = v(1) for a particular interpretation v. Then you infer that v'(A) = v'(1) for an arbitrary interpretation v'. Part of my reason for using the v(.) symbol is to help myself keep the stages of this argument distinct.

Comment author: Vladimir_Nesov 08 October 2010 08:41:38PM 0 points [-]

According to my understanding, you start out knowing that v(A) = v(1) for a particular interpretation v.

If v is an interpretation, it maps (all) terms to elements of corresponding universe, while possible actions are only some formulas, so associated mapping K would map some formulas to the set of actions (which don't have to do anything with any universe). So, we could say that K(1)=1', but K(A) is undefined. K is not an interpretation.

Comment author: Tyrrell_McAllister 08 October 2010 09:30:02PM *  0 points [-]

If v is an interpretation, it maps (all) terms to elements of corresponding universe, while possible actions are only some formulas, . . .

Maybe we're not using the terminology in exactly the same way.

For me, an interpretation of a theory is an ordered pair (D, v), where D is a set (the domain of discourse), and v is a map (the valuation map) satisfying certain conditions. In particular, D is the codomain of v restricted to the constant symbols, so v actually contains everything needed to recover the interpretation. For this reason, I sometimes abuse notation and call v itself the interpretation.

The valuation map v

  • maps constant symbols to elements of D,

  • maps n-ary function symbols to maps from D^n to D,

  • maps n-ary predicate symbols to subsets of D^n,

  • maps sentences of the theory into {T, F}, in a way that satisfies some recursive rules coming from the rules of inference.

Now, in the post, you write

Each such statements defines a possible world Y resulting from a possible action X. X and Y can be thought of as constants, just like A and O, or as formulas that define these constants, so that the moral arguments take the form [X(A) => Y(O)].

(Emphasis added.) I've been working with the bolded option, which I understand to be saying that A and 1 are constant symbols. Hence, given an interpretation (D, v), v(A) and v(1) are elements of D, so we can ask whether they are the same elements.

Comment author: Vladimir_Nesov 08 October 2010 09:53:46PM 0 points [-]

I agree with everything you wrote here...

Comment author: Tyrrell_McAllister 08 October 2010 09:57:21PM *  0 points [-]

I agree with everything you wrote here...

What was your "associated mapping K"? I took it to be what I'm calling the valuation map v. That's the only map that I associate to an interpretation.

Comment author: Vladimir_Nesov 08 October 2010 10:01:04PM *  0 points [-]

K has a very small domain. Say, K("2+2")=K("5")="pull the second lever", K("4") undefined, K("A") undefined. Your v doesn't appear to be similarly restricted.