Lukasiewicz-Slupencki three-valued calculus

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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. :(

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?

(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.