Lukasiewicz-Slupencki three-valued calculus

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  • #1
nomadreid
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(The "L"'s in the two names should have lines through them, sorry).
Slupencki expanded (in 1936) the three-valued Lukasiewicz calculus L3
to L3S in order to make it functionally complete. He did this by adding functor T(.), where T(x) = 1 for all x in {0,1,2}, and two axioms: Tx⇒~Tx and ~Tx⇒Tx. Since val(x)= val(~x) if val(x) = 1, these axioms would seem OK, but what I don't get is why we cannot say then that Tx⇔~Tx poses an unacceptable contradiction. :(
 

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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
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  • #3
nomadreid
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Let me put the question more simply, without reference to a particular system. Can you have a consistent (many-valued) logical system such that there are propositions such that the valuation of the proposition and the valuation of its negation are the same?
 
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TeethWhitener
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(The "L"'s in the two names should have lines through them, sorry).
Slupencki expanded (in 1936) the three-valued Lukasiewicz calculus L3
to L3S in order to make it functionally complete. He did this by adding functor T(.), where T(x) = 1 for all x in {0,1,2}, and two axioms: Tx⇒~Tx and ~Tx⇒Tx. Since val(x)= val(~x) if val(x) = 1, these axioms would seem OK, but what I don't get is why we cannot say then that Tx⇔~Tx poses an unacceptable contradiction. :(
I'm not seeing where the contradiction is. Tx is always 1, therefore ~Tx is always 1 [by the definition of negation val(~x) = 2-x], therefore Tx ⇔ ~Tx is just 1 ⇔ 1.
 
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nomadreid
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Thanks for the reply, TeethWhitener. Start with a tautology, such as T⇔T&T. This becomes (applying T⇔~T) T⇔T&~T. Ditto for getting ~T ⇔T&~T. From T∨~T (and A∨A⇔A), we get T&~T. Contradiction.
Also: If T⇔~T is allowed, then how would one come to the conclusion that Russell's paradox poses a contradiction?
 
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TeethWhitener
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First of all, A&~A is not, in general, a contradiction in 3-valued logic. To see this, plug in the truth value 1 for A (from the set {0, 1, 2}). You get val(~A)=1, val(A&~A)=1. Suspension of the laws of non-contradiction and the excluded middle were part of Lukasiewicz's original motivation for exploring multivalent logics. Secondly, T is not a proposition. T(x) is a unary operator that takes any proposition x and assigns it the truth value of 1.
 
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nomadreid
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Thanks, TeethWhitener. I plead guilty to both counts. I now am clearer on the subject.
 

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