-2

enter image description here

I am currently trying to work on problem 8, but I'm not sure exactly how to start it.

I was thinking of starting it by trying to prove that phi is indeed PL valid. I would have to show how that is connected to superevalution.

I was also thinking of incorporating this:

For any trivalent interpretation I, SVI(phi)=1 I(p)=1 For any precisification C of I, Vc (phi)=1 Therefore Vc(phi)=1

I can also repreat this with indeterminant as well. I wasn't sure if I would need to prove that phi is PL valid first before proving superevaluation.

  • It would help to know which book you're using. You may also want to repeat the relevant definitions of superevaluation interpretations and precisification. – lemontree Feb 19 at 9:20
  • BTW, you are not supposed to prove that phi is valid, but that phi logically follows from a set of premises Gamma. – lemontree Feb 19 at 9:22
  • The book is "Logic for Philosophy" by Theodore Sider – Fiona717 Feb 19 at 15:36
  • Supervaluation is defined: SVI(phi) = 1 if for all C, precicify I, Vc(phi)=1; =0 if for all c, precisify I, Vc(phi)=0, =# if otherwise – Fiona717 Feb 19 at 15:47
  • What are the Cs supposed to represent? Classical models? What is the definition. of 'precisify I'? – sequitur Feb 20 at 17:04
1

The OP lacks a lot of information on how terms used are to be understood. In what follows I'll outline an answer that concerns one standard approach to supervaluation logics, but I'm not sure if that is the one intended by you.

Let a supermodel C be a non-empty set of PL-models (total functions from propositional atoms to {0,1}). Then a supervaluation over C is a function C(.) from atoms to {0, 1/2, 1} with C(A) = 1 iff V(A) = 1, for all members V of C; C(A) = 0 iff V(A) = 0, for all V from C; C(A) = 1/2, otherwise. Finally, A is a superconsequence of set M, iff for every supermodel C, if C(B) = 1, for every B in M, then C(A) =1.

Now for the proof of the claim. Let M be any set of formulas, A a formula and C be any supermodel such that C(B) = 1, for every B in M. Then by definition we have that V(B) = 1, for every V in C. Every member of C is a PL-valuation. Since by hypothesis A is a PL-consequence of M it follows that V(A) = 1, for every V in C. So, A is a superconsequence of M. qed

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.