\documentclass{article} \input linear \input theorem %\usepackage{enumerate} \title{Classical Truth as a Modal Operator} \author{Jason Reed} \def\says{\Longrightarrow} \def\limp{\supset} \def\D{{\cal D}} \def\E{{\cal E}} \def\true{\mathop{\textsf t}\nolimits} \def\false{\mathop\textsf f} \def\cent{{\textbf{\textsf{C}}}} \begin{document} \maketitle We have a judgment $\Gamma \prov \gamma$ where contexts are defined by $$\Gamma ::= \cdot \celse \Gamma, A \celse \Gamma, A\true \celse \Gamma, A\false$$ and propositions by $$A,B,C ::= P \celse \bot \celse \cent A \celse A\limp B$$ and the optionally empty conclusion $\gamma$ is $$\gamma ::= \cdot \celse A$$ Hypotheses $A\true$ and $A\false$ in $\Gamma$ are assumptions of classical truth and falsehood. Bare hypotheses $A$ in $\Gamma$ are intuitionistic assumptions. The right-hand side of the judgment $\Gamma\prov C$ is an intuitionistic conclusion. $\Gamma\prov \cdot$ means that the assumptions in $\Gamma$ are contradictory. Propositions are either propositional variables $P$ or falsehood $\bot$ or constructed from other propositions by the classically true' modality $\cent$ and implication $\limp$. The sequent calculus is as follows. The initial sequents are $$\begin{prooftree} \using init_i \justifies \Gamma, P \prov P \end{prooftree} \qquad \begin{prooftree} \using init_c \justifies \Gamma, P\true, P\false \prov \gamma \end{prooftree} \qquad \begin{prooftree} \using init_{ic} \justifies \Gamma, P, P\false \prov \gamma \end{prooftree}$$ The rules for $\bot$ are $$\begin{prooftree} \using \bot L \justifies \Gamma, \bot \prov \gamma \end{prooftree}$$ $$\begin{prooftree} \using \bot T \justifies \Gamma, \bot \true \prov \gamma \end{prooftree} \qquad \begin{prooftree} \Gamma \prov \gamma \using \bot F \justifies \Gamma, \bot \false \prov \gamma \end{prooftree}$$ The rules for $\cent$ are $$\begin{prooftree} \Gamma, A \true \prov \gamma \using \cent L \justifies \Gamma, \cent A \prov \gamma \end{prooftree} \qquad \begin{prooftree} \Gamma, A \false \prov \cdot \using \cent R \justifies \Gamma \prov \cent A \end{prooftree}$$ $$\begin{prooftree} \Gamma, A \true \prov \gamma \using \cent T \justifies \Gamma, \cent A \true \prov \gamma \end{prooftree} \qquad \begin{prooftree} \Gamma, A \false \prov \gamma \using \cent F \justifies \Gamma, \cent A \false \prov \gamma \end{prooftree}$$ The rules for $\limp$ are $$\begin{prooftree} \Gamma, A\limp B \prov A \qquad \Gamma, B \prov \gamma \using \limp L \justifies \Gamma, A \limp B \prov \gamma \end{prooftree} \qquad \begin{prooftree} \Gamma, A \prov B \using \limp R \justifies \Gamma \prov A\limp B \end{prooftree}$$ $$\begin{prooftree} \Gamma, A \false \prov \gamma_1 \qquad \Gamma,B\true \prov \gamma_2 \using \limp T \justifies \Gamma, A\limp B \true \prov \gamma_1 + \gamma_2 \end{prooftree} \qquad \begin{prooftree} \Gamma, A\true, B\false \prov \gamma \using \limp F \justifies \Gamma, A\limp B \false \prov \gamma \end{prooftree}$$ Where $+$ is defined by $$\cdot + \cdot = \cdot$$ $$C + \cdot = \cdot$$ $$\cdot + C = \cdot$$ where $C_1 + C_2$ is undefined even if $C_1 = C_2$. That is, $\limp T$ is really three rules where one or both (but not neither) of the premises have an empty right-hand side. We may ask, does cut elimination hold for this calculus? Is the rule $$\begin{prooftree} \Gamma\prov A \qquad \Gamma, A \prov \gamma \using cut \justifies \Gamma \prov \gamma \end{prooftree}$$ admissible? In the event that $A$ is of the form $\cent B$ and the derivations of $\Gamma\prov \cent B$ and $\Gamma, \cent B \prov \gamma$ both end with a rule that introduces the connective in the cut formula, the derivation looks like $$\begin{prooftree} $\Gamma, B\false \prov \cdot \using \cent R\justifies \Gamma\prov \cent B$ $\Gamma, B\true \prov \gamma \using \cent L\justifies \Gamma, \cent B \prov \gamma$ \using cut \justifies \Gamma \prov \gamma \end{prooftree}$$ In order to proceed, it seems necessary that we also show the admissibility of a classical cut principle $$\begin{prooftree} \Gamma, B\false \prov \cdot \qquad \Gamma, B\true \prov \gamma \using sync \justifies \Gamma \prov \gamma \end{prooftree}$$ where false assumptions $A\false$ (thought of as continuations, or $A$ readers') synchronize with true assumptions $A\true$ (thought of as `$A$ writers'). But proving this rule admissible is problematic. Consider the left-derivation true-commutative case for $\limp$: $$\begin{prooftree} $\Gamma, A_1 \false, B\false \prov \cdot \qquad \Gamma, A_2\true, B\false \prov \cdot \justifies \Gamma, A_1, \limp A_2\true, B\false \prov \cdot$ \qquad \Gamma, A_1\limp A_2\true, B\true \prov \gamma \using sync \justifies \Gamma, A_1\limp A_2\true \prov \gamma \end{prooftree}$$ We can sync $\Gamma, A_2\true, B\false \prov \cdot$ with $\Gamma, A_1\limp A_2\true, B\true \prov \gamma$ (on $B \true,B\false$) to obtain $\Gamma, A_2\true, A_1\limp A_2\true\prov \gamma$. We can sync $\Gamma, A_1\false, B\false \prov \cdot$ with $\Gamma, A_1\limp A_2\true, B\true \prov \gamma$ (on $B \true,B\false$) to obtain $\Gamma, A_1\false, A_1\limp A_2\true\prov \gamma$. We would like to derive $$\begin{prooftree} \Gamma, A_1 \false , A_1\limp A_2 \true \prov \gamma \qquad \Gamma, A_2 \true A_1\limp A_2 \true \prov \gamma \using \limp T\justifies \Gamma, A_1\limp A_2\true, A_1\limp A_2 \true \prov \gamma \end{prooftree}$$ and apply contraction, but it may not be that $\gamma + \gamma = \gamma$! I still haven't found a concrete counterexample, however. So the question is still open: \begin{question} Are the rules cut and sync admissible? \end{question} \end{document}