logic.tex

\section{Language}

A \emph{language} $\Lang$ consists of a set \textdom{Expr} of \emph{expressions} (metavariable $\expr$), a set \textdom{Val} of \emph{values} (metavariable $\val$), and a set \textdom{State} of \emph{states} (metvariable $\state$) such that
\begin{itemize}
\item There exist functions $\ofval : \textdom{Val} \to \textdom{Expr}$ and $\toval : \textdom{Expr} \pfn \textdom{val}$ (notice the latter is partial), such that
\begin{mathpar} {\All \expr, \val. \toval(\expr) = \val \Ra \ofval(\val) = \expr} \and {\All\val. \toval(\ofval(\val)) = \val} 
\end{mathpar}
\item There exists a \emph{primitive reduction relation} \[(-,- \step -,-,-) \subseteq \textdom{Expr} \times \textdom{State} \times \textdom{Expr} \times \textdom{State} \times (\textdom{Expr} \uplus \set{\bot})\]
  We will write $\expr_1, \state_1 \step \expr_2, \state_2$ for $\expr_1, \state_1 \step \expr_2, \state_2, \bot$. \\
  A reduction $\expr_1, \state_1 \step \expr_2, \state_2, \expr_\f$ indicates that, when $\expr_1$ reduces to $\expr$, a \emph{new thread} $\expr_\f$ is forked off.
\item All values are stuck:
\[ \expr, \_ \step  \_, \_, \_ \Ra \toval(\expr) = \bot \]
\item There is a predicate defining \emph{atomic} expressions satisfying
\let\oldcr\cr
\begin{mathpar}
  {\All\expr. \atomic(\expr) \Ra \toval(\expr) = \bot} \and
  {{
    \begin{inbox}
\All\expr_1, \state_1, \expr_2, \state_2, \expr_\f. \atomic(\expr_1) \land \expr_1, \state_1 \step \expr_2, \state_2, \expr_\f \Ra {}\\\qquad\qquad\qquad\quad~~ \Exists \val_2. \toval(\expr_2) = \val_2
    \end{inbox}
}}
\end{mathpar}
In other words, atomic expression \emph{reduce in one step to a value}.
It does not matter whether they fork off an arbitrary expression.
\end{itemize}

\begin{defn}
  An expression $\expr$ and state $\state$ are \emph{reducible} (written $\red(\expr, \state)$) if
  \[ \Exists \expr_2, \state_2, \expr_\f. \expr,\state \step \expr_2,\state_2,\expr_\f \]
\end{defn}

\begin{defn}[Context]
  A function $\lctx : \textdom{Expr} \to \textdom{Expr}$ is a \emph{context} if the following conditions are satisfied:
  \begin{enumerate}[itemsep=0pt]
  \item $\lctx$ does not turn non-values into values:\\
    $\All\expr. \toval(\expr) = \bot \Ra \toval(\lctx(\expr)) = \bot $
  \item One can perform reductions below $\lctx$:\\
    $\All \expr_1, \state_1, \expr_2, \state_2, \expr_\f. \expr_1, \state_1 \step \expr_2,\state_2,\expr_\f \Ra \lctx(\expr_1), \state_1 \step \lctx(\expr_2),\state_2,\expr_\f $
  \item Reductions stay below $\lctx$ until there is a value in the hole:\\
    $\All \expr_1', \state_1, \expr_2, \state_2, \expr_\f. \toval(\expr_1') = \bot \land \lctx(\expr_1'), \state_1 \step \expr_2,\state_2,\expr_\f \Ra \Exists\expr_2'. \expr_2 = \lctx(\expr_2') \land \expr_1', \state_1 \step \expr_2',\state_2,\expr_\f $
  \end{enumerate}
\end{defn}

\subsection{Concurrent language}

For any language $\Lang$, we define the corresponding thread-pool semantics.

\paragraph{Machine syntax}
\[
	\tpool \in \textdom{ThreadPool} \eqdef \bigcup_n \textdom{Exp}^n
\]

\judgment[Machine reduction]{\cfg{\tpool}{\state} \step
  \cfg{\tpool'}{\state'}}
\begin{mathpar}
\infer
  {\expr_1, \state_1 \step \expr_2, \state_2, \expr_\f \and \expr_\f \neq \bot}
  {\cfg{\tpool \dplus [\expr_1] \dplus \tpool'}{\state} \step
     \cfg{\tpool \dplus [\expr_2] \dplus \tpool' \dplus [\expr_\f]}{\state'}}
\and\infer
  {\expr_1, \state_1 \step \expr_2, \state_2}
  {\cfg{\tpool \dplus [\expr_1] \dplus \tpool'}{\state} \step
     \cfg{\tpool \dplus [\expr_2] \dplus \tpool'}{\state'}}
\end{mathpar}

\clearpage
\section{Logic}

To instantiate Iris, you need to define the following parameters:
\begin{itemize}