thys/notes.tex
author Christian Urban <christian dot urban at kcl dot ac dot uk>
Fri, 30 Jan 2015 11:42:53 +0000
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updated
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\documentclass[11pt]{article}
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\begin{document}
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\noindent
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A lemma which might be true, but can also be false, is as follows:
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\begin{center}
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\begin{tabular}{lll}
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If   & (1) & $v_1 \succ_{der\;c\;r} v_2$,\\
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     & (2) & $\vdash v_1 : der\;c\;r$, and\\ 
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     & (3) & $\vdash v_2 : der\;c\;r$ holds,\\
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then &     & $inj\;r\;c\;v_1 \succ_r inj\;r\;c\;v_2$ also holds.  
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\end{tabular}
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\end{center}
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\noindent It essentially states that if one value $v_1$ is 
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bigger than $v_2$ then this ordering is preserved under 
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injections. This is proved by induction (on the definition of 
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$der$\ldots this is very similar to an induction on $r$).
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\bigskip
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\noindent
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The case that is still unproved is the sequence case where we 
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assume $r = r_1\cdot r_2$ and also $r_1$ being nullable.
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The derivative $der\;c\;r$ is then
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\begin{center}
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$der\;c\;r = ((der\;c\;r_1) \cdot r_2) + (der\;c\;r_2)$
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\end{center}
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\noindent 
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or without the parentheses
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\begin{center}
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$der\;c\;r = (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$
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\end{center}
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\noindent 
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In this case the assumptions are
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\begin{center}
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\begin{tabular}{ll}
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(a) & $v_1 \succ_{(der\;c\;r_1) \cdot r_2 + der\;c\;r_2} v_2$\\
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(b) & $\vdash v_1 : (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$\\
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(c) & $\vdash v_2 : (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$\\
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(d) & $nullable(r_1)$
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\end{tabular}
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\end{center}
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\noindent 
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The induction hypotheses are
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\begin{center}
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\begin{tabular}{ll}
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(IH1) & $\forall v_1 v_2.\;v_1 \succ_{der\;c\;r_1} v_2
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\;\wedge\; \vdash v_1 : der\;c\;r_1 \;\wedge\; 
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\vdash v_2 : der\;c\;r_1\qquad$\\
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      & $\hfill\longrightarrow
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         inj\;r_1\;c\;v_1 \succ{r_1} \;inj\;r_1\;c\;v_2$\smallskip\\
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(IH2) & $\forall v_1 v_2.\;v_1 \succ_{der\;c\;r_2} v_2
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\;\wedge\; \vdash v_2 : der\;c\;r_2 \;\wedge\; 
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\vdash v_2 : der\;c\;r_2\qquad$\\
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      & $\hfill\longrightarrow
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         inj\;r_2\;c\;v_1 \succ{r_2} \;inj\;r_2\;c\;v_2$\\
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\end{tabular}
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\end{center}
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\noindent 
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The goal is
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\[
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(goal)\qquad
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inj\; (r_1 \cdot r_2)\;c\;v_1 \succ_{r_1 \cdot r_2} 
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inj\; (r_1 \cdot r_2)\;c\;v_2
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\]
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\noindent 
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If we analyse how (a) could have arisen (that is make a case
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distinction), then we will find four cases:
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\begin{center}
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\begin{tabular}{ll}
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LL & $v_1 = Left(w_1)$, $v_2 = Left(w_2)$\\
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LR & $v_1 = Left(w_1)$, $v_2 = Right(w_2)$\\
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RL & $v_1 = Right(w_1)$, $v_2 = Left(w_2)$\\
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RR & $v_1 = Right(w_1)$, $v_2 = Right(w_2)$\\
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\end{tabular}
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\end{center}
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\noindent 
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We have to establish our goal in all four cases. 
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\subsubsection*{Case LR}
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The corresponding rule (instantiated) is:
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\begin{center}
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\begin{tabular}{c}
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$len\,|w_1| \geq len\,|w_2|$\\
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\hline
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$Left(w_1) \succ_{(der\;c\;r_1) \cdot r_2 + der\;c\;r_2} Right(w_2)$
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\end{tabular}
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\end{center}
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\noindent 
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This means we can also assume in this case
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\[
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(e)\quad len\,|w_1| \geq len\,|w_2|
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\] 
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\noindent 
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which is the premise of the rule above.
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Instantiating $v_1$ and $v_2$ in the assumptions (b) and (c)
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gives us
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\begin{center}
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\begin{tabular}{ll}
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(b*) & $\vdash Left(w_1) : (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$\\
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(c*) & $\vdash Right(w_2) : (der\;c\;r_1) \cdot r_2 + der\;c\;r_2$\\
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\end{tabular}
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\end{center}
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\noindent Since these are assumptions, we can further analyse
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how they could have arisen according to the rules of $\vdash
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\_ : \_\,$. This gives us two new assumptions
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\begin{center}
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\begin{tabular}{ll}
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(b**) & $\vdash w_1 : (der\;c\;r_1) \cdot r_2$\\
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(c**) & $\vdash w_2 : der\;c\;r_2$\\
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\end{tabular}
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\end{center}
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\noindent 
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Looking at (b**) we can further analyse how this
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judgement could have arisen. This tells us that $w_1$
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must have been a sequence, say $u_1\cdot u_2$, with
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   143
\begin{center}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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\begin{tabular}{ll}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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(b***) & $\vdash u_1 : der\;c\;r_1$\\
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   146
       & $\vdash u_2 : r_2$\\
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   147
\end{tabular}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   148
\end{center}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   149
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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\noindent 
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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Instantiating the goal means we need to prove
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   152
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   153
\[
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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inj\; (r_1 \cdot r_2)\;c\;(Left(u_1\cdot u_2)) \succ_{r_1 \cdot r_2} 
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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inj\; (r_1 \cdot r_2)\;c\;(Right(w_2))
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   156
\]
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   157
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   158
\noindent 
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   159
We can simplify this according to the rules of $inj$:
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   160
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   161
\[
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   162
(inj\; r_1\;c\;u_1)\cdot u_2 \succ_{r_1 \cdot r_2} 
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   163
(mkeps\;r_1) \cdot (inj\; r_2\;c\;w_2)
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   164
\]
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   165
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   166
\noindent
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   167
This is what we need to prove.
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
diff changeset
   168
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents: 55
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   169
55
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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   170
\subsubsection*{Case RL}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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The corresponding rule (instantiated) is:
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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\begin{center}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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\begin{tabular}{c}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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$len\,|w_1| > len\,|w_2|$\\
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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\hline
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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$Right(w_1) \succ_{(der\;c\;r_1) \cdot r_2 + der\;c\;r_2} Left(w_2)$
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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\end{tabular}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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\end{center}
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Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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   181
c33cfa1e813a added some notes (still incomplete)
Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
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   182
c33cfa1e813a added some notes (still incomplete)
Christian Urban <christian dot urban at kcl dot ac dot uk>
parents:
diff changeset
   183
\end{document}