For any finite valued logic, where we have some connective "C" where detachment is a rule of inference for that connective, and we have truth tables, we can basically turn what the truth tables tell us into an axiomization of the logic using functioral variables (we'll also want |=Cpp to hold, otherwise we'll run into a problem with the way the transformation works).
With three-valued conditional-negation Lukasiewicz logic we have the following tables:
C 0 1 2 N
0 2 2 2 2
1 1 2 2 1
2* 0 1 2 0
The table tells us that from an algebraic perspective, we have the following equations: C00=2, C01=2, C02=2, C10=1, C11=2, C12=2, C20=0, C21=1, C22=2, N0=2, N1=1, N2=0. It turns out that if x=y, then C$\delta$x$\delta$y, where $\delta$ is a variable functor of one argument. So, to make an axiom set for a finite-valued logic first suppose we find each of these C$\delta$x$\delta$y for the finite-valued logic under study. Make each of those C$\delta$x$\delta$y into an axiom. Next make the designated value into an axiom. Lastly, for a finite-valued logic where it's values belong to the set {a, ..., o} make the following into an axiom:
C$\delta$aC$\delta$b...C$\delta$o$\delta$p. When you've done those three steps you have a complete and sound axiomization for the logical system.
So for the four-valued C-N Lukasiewicz logic with the following table:
C 1 2 3 0 N
1* 1 2 3 0 0
2 1 1 3 3 3
3 1 2 1 2 2
0 1 1 1 1 1
the following is a complete and sound axiomatization under the rules of substitution for propositional variables, substitution for the unary functorial variable, and detachment:
- 1
- C$\delta$1$\delta$C11.
- C$\delta$2$\delta$C12.
- C$\delta$3$\delta$C13.
- C$\delta$0$\delta$C10.
- C$\delta$1$\delta$C21.
- C$\delta$1$\delta$C22.
- C$\delta$3$\delta$C23.
- C$\delta$3$\delta$C20.
- C$\delta$1$\delta$C31.
- C$\delta$2$\delta$C32.
- C$\delta$1$\delta$C33.
- C$\delta$2$\delta$C30.
- C$\delta$1$\delta$C01.
- C$\delta$1$\delta$C02.
- C$\delta$1$\delta$C03.
- C$\delta$1$\delta$C00.
- C$\delta$0$\delta$N1.
- C$\delta$3$\delta$N2.
- C$\delta$2$\delta$N3.
- C$\delta$1$\delta$N0.
- C$\delta$1C$\delta$2C$\delta$3C$\delta$0$\delta$p.