- #1

nomadreid

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Slupencki expanded (in 1936) the three-valued Lukasiewicz calculus L

_{3}

to L

_{3}S 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. :(