Thoughts on sigma-separation and a good sufficient condition

diff --git a/ac_notes.tex b/ac_notes.tex
index 498812eab9a78b14566e9f0de8fc43d4fb361995..d8ff891551cc1c7b5efa502e00e23dc437ab69cb 100644 (file)
--- a/ac_notes.tex
+++ b/ac_notes.tex
@@ -13,12 +13,12 @@
 \r
 \maketitle\r
 \r
-\newcommand{\Prgm}[2]{[#1\Comma #2]}\r
+\renewcommand{\Lam}[2]{\lambda#1.\, \{\!\!\{#2\}\!\!\}}\r
 \r
 \subparagraph{Syntax}\r
 \[\begin{array}{lll}\r
-  \tm, \tmtwo & \ddef & \var \mid \tm\,\tmtwo \mid \Lam\var \Prgm{\tm}{\vec\tm} \\\r
-  n & \ddef & \Lam\var \Prgm{n}{\vec n} \mid \var\,\vec n \\\r
+  \tm, \tmtwo & \ddef & \var \mid \tm\,\tmtwo \mid \Lam\var{\tm\Comma\vec\tm} \\\r
+  n & \ddef & \Lam\var{n\Comma\vec n} \mid \var\,\vec n \\\r
   \\\r
   C & \ddef & \Box \mid C\,\tm \mid \tm\,C \\\r
   P & \ddef & \vec\tm \Comma \Box \Comma \vec\tm \mid \vec\tm \Comma C[\Lam\var P] \Comma \vec\tm  \\\r
@@ -26,9 +26,9 @@
 \r
 \subparagraph{Reduction}\r
 \[\begin{array}{lll}\r
-  P[C[(\Lam\var \Prgm\tm{\vec\tm})\,\tmtwo]] & \Red{}{\var\in \Prgm\tm{\vec\tm}} &\r
+  P[C[(\Lam\var{\tm\Comma\vec\tm})\,\tmtwo]] & \Red{}{\var\in \tm,\vec\tm} &\r
    P[C[\tm\Subst\var\tmtwo]\Comma \vec\tm\Subst\var\tmtwo]\\\r
-   P[C[(\Lam\var \Prgm\tm{\vec\tm})\,\tmtwo]] & \Red{}{\var\not\in\Prgm\tm{\vec\tm}} &\r
+   P[C[(\Lam\var{\tm\Comma\vec\tm})\,\tmtwo]] & \Red{}{\var\not\in\tm,\vec\tm} &\r
     P[C[\tm] \Comma \vec\tm\Comma\tmtwo]\\\r
 \end{array}\]\r
 \r
@@ -38,6 +38,77 @@
   \item Ogni strategia e' perpetua!\r
 \end{itemize}\r
 \r
+\clearpage\r
+\[ E ::= \Box \mid E\, t \mid t\, E \mid \Lam\var{\vec t\Comma E \Comma \vec t} \]\r
+\r
+\begin{definition}[Unlockable variable]\r
+  A variable $\var$ is unlockable in a context $E$ if:\r
+  \begin{itemize}\r
+    \item it is not bound in $E$, or\r
+    \item $E[\cdot] = E'[\vartwo\, \vec\tm \, (\Lam{\cdots\var\cdots}{\vec\tmtwo \Comma E'[\cdot]\Comma\vec\tmthree})]$\r
+     and $\vartwo$ is unlockable in $E'$.\r
+  \end{itemize}\r
+\end{definition}\r
+\r
+\begin{lemma}[Elimination of unlockable variables]\r
+  For every set of terms $\tm_1, \ldots, \tm_n$\r
+  there exist terms $\tm'_1, \ldots, \tm'_n$\r
+  without unlockable variables and\r
+  such that $\tm_i$ is unseparable from $\tm'_i$.\r
+\end{lemma}\r
+\r
+{\color{red}\r
+ Non esattamente!\r
+ I termini hanno gli stessi path, e le variabili unlockable\r
+ del secondo sono iin path virtuali nel primo\r
+\r
+ Piu' esattamente, hai portato il garbage che vuoi far divergere\r
+  al top level.\r
+}\r
+\r
+\begin{definition}\r
+  Transformation removing an unlockable variable bound at position $\pi$:\r
+  \[\tau_{n::\pi}[\alpha] := \lambda x_1..x_n\,x.\, \alpha\,\vec x\,(\tau_\pi[x])\]\r
+\end{definition}\r
+\r
+{\color{red}\r
+Warning! It creates unlockable variables, but in positions whose paths were\r
+previously virtual}\r
+\r
+\begin{lemma}\r
+  For every semi-$\sigma$-separable $\tm$ and $\tmtwo$,\r
+   there exist $\sigma$ s.t $\tm\sigma = \ldots \Comma C[i] \Comma \ldots$\r
+   and $i \not\preceq \tmtwo\sigma$ (and therefore, $\tm\sigma$ and $\tmtwo\sigma$\r
+   are semi-$\sigma$-separable by ???).\r
+\end{lemma}\r
+\r
+\begin{theorem}\r
+  $\tm$ and $\tmtwo$ are semi-$\sigma$-separable if \ldots\r
+\end{theorem}\r
+\r
+\clearpage\r
+\newcommand{\HeadOf}[1]{\operatorname{hd}(#1)}\r
+\newcommand{\ArityOf}[1]{\operatorname{len}(#1)}\r
+\begin{itemize}\r
+  \item \textbf{A sufficient condition for separability:}\r
+  \item \textbf{Goal.} semi-$\sigma$-separating $\tm$ from $\tmtwo$ (both inerts with pacmans/$\Omega$).\r
+  \item Let $U \defeq \{\tmtwo' \mid \tmtwo'\preceq\tmtwo \land \HeadOf{\tmtwo'}=\HeadOf{\tm}\r
+  \land \ArityOf{\tmtwo'} \leq \ArityOf{\tm}\} = \{u_1, \ldots, u_N\}$.\r
+  \item \textbf{Theorem.} The problem is separable if there exist $\pi_1\ldots\pi_N$\r
+  $\eta$-difference paths between\r
+  $\tm$ and $\tmtwo_1\ldots\tmtwo_N$ that are pairwise compatible on $\tm$ (note: also between themselves).\r
+  \item \textbf{Definition ($\eta$-difference path).}\r
+  \item \textbf{Definition ($\sqsubset$).} Initial fragments of paths\r
+   \item \textbf{Definition (Compatible paths).} ``$\pi_j$ is compatible with $\pi_i$ on $\tm$'' iff:\r
+    \[\forall\pi_i'\sqsubseteq\pi_i\text{ s.t. }\HeadOf{\pi_i'[t]} = \HeadOf{t}\r
+    \text{, then }\pi_j[\pi_i'[t]] \neq \Omega.\]\r
+  \item Resta da dimostrare che esista un modo di pre-processare i termini nostri\r
+   e ottenere degli inerti scelti.\r
+\end{itemize}\r
+\r
+\r
+\r
+\r
 \clearpage\r
 \section*{Syntactic Condition}\r
 \r