Modal Logic and Provability: Philosophy Forums

Modal Logic and Provability

taiho
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Posted 05/08/08 - 02:25 PM: Subject: Modal Logic and Provability	#1
I'm having a bit of trouble understanding some applications of modal logic to provability. Any explanation or help on these would be helpful. For reference BewPA means provable in PA. Take a "world" to be a model of PA and stipulate that a world A has access to a world B iff whenever BewPA(P) is true in A, P is true in B. 1. Show that a sentence BewPA(P) is true in a world A iff P is true in every world accessible from A. Going left to right (assuming BewPA is true in A), it seems to follow by definition of the accessibility relation that if BewPA(P) is true in a world A, P is true in every world accessible. I'm having a bit of trouble going right to left - If P is true in every world accessible from A, BewPA(P) is true in world A. I think the right strategy is to try to prove the contrapositive: If NOT BewPA(P) is true in world A, there is a world accessible from A where NOT P is true, but I'm not sure how to do this. Any help would be appreciated! Thank you!

MoeBlee
aka I. Kabruob

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Posted 05/08/08 - 06:03 PM:	#2
taiho wrote: I'm having a bit of trouble understanding some applications of modal logic to provability. Any explanation or help on these would be helpful. For reference BewPA means provable in PA. I.e., BewPA(P) iff PA \|- P. taiho wrote: Take a "world" to be a model of PA and stipulate that a world A has access to a world B iff whenever BewPA(P) is true in A, P is true in B. I've not worked with this kind of thing very much, but are you quite sure of the above definition? Unpacking it, reads like this: A world A has acesss to a world B iff for all sentences P, if "PA proves P" is true in A then P is true in B. Are you sure that's right?

7
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Posted 05/08/08 - 10:28 PM:	#3
I think the right strategy is to try to prove the contrapositive: If NOT BewPA(P) is true in world A, there is a world accessible from A where NOT P is true, but I'm not sure how to do this. I would guess that Bew works like the necessity operator. Therefore, if Bew(P) is true at w and wRv, then P is true at v and if ~Bew(P) is true at w then for some v such that wRv, ~P is true at v. Think of it like necessity. If P is not necessary at w, then there is some accessible world for which P is false. Edited by 7 on 05/08/08 - 10:34 PM

muxol
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Posted 05/09/08 - 02:08 PM:	#4
I don't get the exercise. It is asking you to "prove" a definition, viz. that given for the modal operator Bew. Well, the proof is trivial. Just cite the semantic definition of Bew. Notice that the worlds are partially ordered by inclusion, since each world is an extension of PA (in the intended sense you give). I'm not sure that makes a difference, since it appears your exercise is misstated.

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