Lukasiewicz-Slupencki three-valued calculus

[nomadreid]

(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. :(

 

[nomadreid]

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?

 

[TeethWhitener]

(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.

 

[nomadreid]

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?

 

[TeethWhitener]

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.

 

[nomadreid]

Thanks, TeethWhitener. I plead guilty to both counts. I now am clearer on the subject.