# Lukasiewicz-Slupencki three-valued calculus

Gold Member
(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. :(

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?

• Gold Member
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
Gold Member
(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.

• Gold Member
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
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