SIXTEEN$_3$ in Light of Routley Stars

Abstract

For one of the most well-known many-valued logics FDE, there are several semantics, including the star semantics by Richard Routley and Valerie Routley, the two-valued relational semantics by Michael Dunn and the four-valued semantics by Nuel Belnap. The last semantics inspired Yaroslav Shramko and Heinrich Wansing to introduce the trilattice SIXTEEN$_3$. In this article, we offer two alternative semantical presentations for SIXTEEN$_3$, by applying the Routleys’ semantics and the Dunn semantics. Based on our new semantics, we discuss related systems with less truth values, as well as the relation to FDE-based modal logics.

The work reported in this paper started during DS’s visit to Japan which was supported by JSPS KAKENHI Grant Number JP18K12183 granted to HO. HO was supported by a Sofja Kovalevskaja Award of the Alexander von Humboldt-Foundation, funded by the German Ministry for Education and Research. The work of DS has been carried out as part of the research project “FDE-based modal logics”, supported by the Deutsche Forschungsgemeinschaft, DFG, grant WA 936/13-1. We would like to thank Sergei Odintsov, Yaroslav Shramko, Heinrich Wansing, Zach Weber and the referees for helpful discussions and/or comments on an earlier draft.

Appendices

A Details of the Proof of Theorem 3

We prove the contrapositive. Assume $A\not \vdash B$. Then, by Lindenbaum’s lemma, there is a prime theory $\varGamma$ such that $A\in \varGamma$ and $B\not \in \varGamma$. We then define a two-star interpretation $\langle W, g, *_1, *_2, v \rangle$ as follows:

If we can show that the above condition holds for all formulas, then the result follows since at $a\in W$, $I(a, A)=1$ but $I(a, B)\ne 1$, i.e. $A\not \models _*\,B$. We prove this by induction on the complexity of A. We only prove the cases for ${{\sim }}$ and ${{\sim }}_f$, since the cases for $\wedge$ and $\vee$ are straightforward.

Case 1. If A is an element of $\mathsf {Prop}$, the result holds by definition.

Case 2. If $A={{\sim }}B$, then

Case 3. If $A={{\sim }}_f B$, then

This completes the proof.   $\square$

B Details of the Proof of Proposition 4

We prove the contrapositive. Assume $A\not \models _r\,B$. Then, there is a one-star interpretation $\langle W, g, *, r \rangle$ such that $Ar_g1$, but not $Br_g1$. We then define a two-star interpretation $\langle W, g *_1, *_2, v \rangle$ as follows:

If we can show that the above condition holds for all formulas, then the result follows since at $a\in W$, $v(a, A)=1$ but $v(a, B)\ne 1$, i.e. $A\not \models _{*,g}\,B$. We prove this by induction. We only prove the cases for ${{\sim }}$ and ${{\sim }}_f$, since the cases for $\wedge$ and $\vee$ are straightforward.

Case 1. If A is an element of $\mathsf {Prop}$, the result holds by definition.

Case 2. If $A={{\sim }}B$, then

Case 3. If $A={{\sim }}_f B$, then

This completes the proof.   $\square$

C Details of the Proof for Proposition 5

We prove the contrapositive. Assume $A\not \models _{*,g}\,B$. Then, there is a two-star interpretation $\langle W, g, *_1, *_2, v \rangle$ such that $I(g, A)=1$ but $I(g, B)\ne 1$. We then define a one-star interpretation $\langle W, g, *, r \rangle$ as follows:

If we can show that the above condition holds for all formulas, then the result follows since at $a\in W$, $Ar_a1$ but not $Br_a1$, i.e. $A\not \models _r\,B$. We prove this by induction. We only prove the cases for ${{\sim }}$ and ${{\sim }}_f$, since the cases for $\wedge$ and $\vee$ are straightforward.

Case 1. If A is an element of $\mathsf {Prop}$, the result holds by definition.

Case 2. If $A={{\sim }}B$, then

Case 3. If $A={{\sim }}_f B$, then

This completes the proof.   $\square$