%%% This is the scribe notes template for CS611 %%% There are several comments preceded by CS611: and boxed in %%%%'s %%% which indicate where macros should be altered to set up the header %%% for the paper. Your Notes should go at the comment SCRIBE NOTES GO HERE!. %%% In the various .sty files that accompany this .tex file you will %%% find LaTeX macros that make it easier to typeset inference rules %%% and programming language constructs. You must make sure that the %%% file proof.sty is in a path searched by LaTeX when you try to %%% use this file. Take a look to see what macros are available--it %%% will save you time and make the notes look better. Feel free to %%% extend the set of macros--post them to the newsgroup and contact %%% the course staff if you come up with some good ones so they can be %%% added to the template. %%% This template includes examples of how to use some of the macros %%% to give you an idea of how they work. (Delete the examples when %%% you do your scribing.) \documentclass{article} \usepackage{611-lecture} %\usepackage{graphicx} \usepackage{amsmath,amssymb,amsthm,amsfonts} %\usepackage[all]{xy} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% CS611: Please fill in these macros as appropriate: \lecture{20} %% Lecture number \title{Domain Constructions} %% Title of lecture %\author{Michael O'Connor} %% name of scribe \date{16 October 2006} %% Date of lecture, e.g., 1 January 2001 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % See 611.sty for a variety of macros that will be helpful in % typesetting the lecture. Here are a few of particular interest: % % "x" x in keyword font (e.g., "if", "#t") % _x_ x in italics % \nm{n} n in slanted font (used for abbreviations) % e in angle brackets % \lt less-than sign % \gt greater-than sign % \SB{x} x in semantic brackets % \Tr x{y} x[[y]] with x in calligraphic font % (if x is more than a single character, use \Tr{x}{y}) \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newcommand{\LET}{\impfnt{let}~} \newcommand{\IN}{~\impfnt{in}~} \newcommand{\ENV}{~\impfnt{Env}~} \newcommand{\FENV}{~\impfnt{FEnv}~} \newcommand{\Z}{\mathbb{Z}} \newcommand{\group}[1]{\left\langle{#1}\right\rangle} \theoremstyle{definition} \newtheorem*{defn}{Definition} \newcommand{\nondet}{\left[\!\kern1pt\right]} \renewcommand\phi\varphi \renewcommand\wp[2]{\mathsf{wp}~{#1}~{#2}} \newcommand\wlp[2]{\mathsf{wlp}~{#1}~{#2}} \renewcommand\({\begin{eqnarray*}} \renewcommand\){\end{eqnarray*}} \newcommand\LOOKUP[2]{\mathrm{LOOKUP}~{#1}~{#2}} \newcommand\UPDATE[3]{\mathrm{UPDATE}~{#1}~{#2}~{#3}} \newcommand\MALLOC[2]{\mathrm{MALLOC}~{#1}~{#2}} \newcommand\EMPTY{\mathrm{EMPTY\mbox{-}STORE}} \renewcommand\dom[1]{\mathrm{dom}\,{#1}} \newcommand\p[2]{\langle{#1},\,{#2}\rangle} \newcommand\bigcdot{\mathrel{\raisebox{1pt}{$\scriptscriptstyle\bullet$}}} \newcommand\holed[1]{[\,#1\,]} \newcommand\hole{\holed\bigcdot} \newcommand\context[1]{E\kern1pt\holed{#1}} \newcommand\contextHole{\context\bigcdot} \newcommand\goesto[2]{\underset{#2}{\overset{#1}\longrightarrow}} \newcommand\ifthenelse[3]{\mathsf{if\ }#1\mathsf{\ then\ }#2\mathsf{\ else\ }#3} \newcommand\whiledo[2]{\mathsf{while\ }#1\mathsf{\ do\ }#2} \newcommand\letin[3]{\mathsf{let\ }#1 = #2\mathsf{\ in\ }#3} \newcommand\letrec[5]{\mathsf{letrec\ }#1 = #2\mathsf{\ and\ \ldots\ and\ }#3 = #4\mathsf{\ in\ }#5} \newcommand\letrecone[3]{\mathsf{letrec\ }#1 = #2\mathsf{\ in\ }#3} \newcommand\true{\ensuremath{\mathsf{true}}} \newcommand\false{\ensuremath{\mathsf{false}}} \newcommand\error{\ensuremath{\mathsf{error}}} \newcommand\pca[3]{\{#1\}\kern1pt{#2}\kern1pt\{#3\}} \newcommand\states{\Set{St}} \newcommand\rtc{^{\textstyle *}} \newcommand\sat\vDash \newcommand\force\vdash \newcommand\hyphen{\mbox{-}} \newcommand\lookup[2]{\nm{LOOKUP}~#1~\mquote{#2}} \newcommand\update[3]{\nm{UPDATE}~#1~\mquote{#2}~#3} \newcommand\SBk[1]{\SB{#1}k} \newcommand\fix[1]{\mathsf{fix}\,{#1}} \newlength\reasonwidth \setlength\reasonwidth{3cm} \newcommand\reasoning[1]{\def\longest{#1}\settowidth{\reasonwidth}{$\displaystyle\longest$}\addtolength{\reasonwidth}{5mm}} %dck 2/12/98 \newcommand\reason[2]{\makebox[\reasonwidth][l]{$\displaystyle{#1}$}\mbox{#2}} \renewcommand\inj[1]{\mathsf{in}_{#1}} \newcommand\proj[1]{\pi_{#1}} \newcommand{\dlt}{\sqsubseteq} \newcommand\floor[1]{\lfloor{#1}\rfloor} \newcommand\cf[1]{[\kern1pt{#1}\kern1pt]} \begin{document} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% CS611: SCRIBE NOTES GO HERE! %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Reprise} Last time, guided by the intuition that the programs $\whiledo bc$ and $\ifthenelse b{c;\whiledo bc}{"skip"}$ should be equivalent, we defined the denotation of the statement $\whiledo bc$ as the least solution to the equation \( \mathcal W &\definedas& {\lam{\sigma\in\Sigma}{ \begin{cases} (\mathcal W)^*(\Tr Cc\sigma), & \mbox{if $\Tr Bb\sigma$,}\\ \sigma, & \mbox{otherwise}\\ \end{cases} }} \) in $\Sigma\to\Sigma_\bot$; that is, the least fixpoint of the operator \( F &\definedas& \lam{w\in\Sigma\to\Sigma_\bot}{\lam{\sigma\in\Sigma}{ \begin{cases} (w)^*(\Tr Cc\sigma), & \mbox{if $\Tr Bb\sigma$,}\\ \sigma, & \mbox{otherwise}\\ \end{cases} }} \) of type $(\Sigma\to\Sigma_\bot) \to (\Sigma\to\Sigma_\bot)$. More simply, we might write \( F &\definedas& \lam{w\in\Sigma\to\Sigma_\bot}{\lam{\sigma\in\Sigma}{\ifthenelse{\Tr Bb\sigma}{(w)^*(\Tr Cc\sigma)}\sigma}} \) with the understanding that the if-then-else here is purely mathematical. Here if $w:\Sigma\to\Sigma_\bot$, then $(w)^*:\Sigma_\bot\to\Sigma_\bot$ is the _lift_ of $w$, which sends $\bot$ to $\bot$ and $x$ to $w(x)$ for $x\in\Sigma-\{\bot\}$. In order to show that the least fixpoint of $F$ exists, we will apply the Knaster--Tarski theorem. However, we only proved the Knaster--Tarski theorem for the partial order of subsets of some universal set ordered by set inclusion $\subseteq$. We need to extend it to the more general case of chain-complete partial orders (CPOs). To apply this theorem, we must know that the function space $\Sigma\to\Sigma_\bot$ is a CPO and that $F$ is a continuous map on this space. \section{Chain-Complete Partial Orders and Continuous Functions} Recall that a binary relation $\dlt$ on a set $X$ is a _partial order_ if it is \begin{itemize} \item _reflexive_: $x\dlt x$ for all $x\in X$; \item _transitive_: for all $x,y,z\in X$, if $x\dlt y$ and $y\dlt z$, then $x\dlt z$; \item _antisymmetric_: for all $x,y\in X$, if $x\dlt y$ and $y\dlt x$, then $x=y$. \end{itemize} It is a _total order_ if for all $x,y\in X$, either $x\dlt y$ or $y\dlt x$. If $A\subseteq X$, we say that $x$ is an _upper bound_ for $A$ if $y\dlt x$ for all $y\in A$. We say that $x$ is a _least upper bound_ or _supremum_ of $A$ if $x$ is an upper bound for $A$, and for all other upper bounds $y$ of $A$, $x\dlt y$. Upper bounds and suprema need not exist. For example, the set of natural numbers $\mathbb N$ under its natural order $\leq$ has no supremum in $\mathbb N$. However, if the supremum of any set exists, it is unique. A partially ordered set is said to be _complete_ if all subsets have suprema. The supremum of a set $C$, if it exists, is denoted $\bigsqcup C$. Note that all elements of $X$ are (vacuously) upper bounds of the empty set $\varnothing$, so if the supremum of $\varnothing$ exists, then it is necessarily the least element of the entire set. In this case we give it the name $\bot$. A _chain_ is a subset of $X$ that is totally ordered by $\dlt$. For example, in the partial order of subsets of $\{0,1,2\}$ ordered by set inclusion, the set $\{\varnothing,\{2\},\{1,2\},\{0,1,2\}\}$ is a chain. A partially ordered set is _chain-complete_ if all nonempty chains have suprema. A chain-complete partially ordered set is called a CPO. The empty chain $\varnothing$ is not included in the definition of chain-complete, but if the empty chain also has a supremum, then it is necessarily the least element $\bot$ of the CPO. A CPO with a least element $\bot$ is called _pointed_. Let $X$ and $Y$ be CPOs (we'll use $\dlt$ to denote the partial order in both $X$ and $Y$). A function $f:X\to Y$ is _monotone_ if $f$ preserves order; that is, for all $x,y\in X$, if $x\dlt y$ then $f(x)\dlt f(y)$. For example, the exponential function $\lam x{e^x}:\mathbb{R}\to\mathbb{R}$ is monotone. A function $f:X\to Y$ is _continuous_ if $f$ preserves suprema of nonempty chains; that is, if $C\subseteq X$ is a nonempty chain in $X$, then $\bigsqcup_{x\in C} f(x)$ exists and equals $f(\bigsqcup C)$. Here $\bigsqcup_{x\in C} f(x)$ is alternate notation for $\bigsqcup\,\{f(x)\mid x\in C\}$. Every continuous map is monotone: if $x\dlt y$, then $y = \bigsqcup\{x,y\}$, so by continuity $f(y) = f(\bigsqcup\{x,y\}) = \bigsqcup\,\{f(x),f(y)\}$, which implies that $f(x)\dlt f(y)$. In the definition of continuity, we excluded the empty chain $\varnothing$. If it were included, then a continuous function would have to preserve $\bot$; that is, $f(\bot)=\bot$. A continuous function that satisfies this property is called _strict_. We do not include $\varnothing$ in the definition of continuous functions, because we wish to consider non-strict functions, such as the $F$ of section 1. \section{The Knaster--Tarski Theorem in CPOs} Let $F:D\to D$ be any continuous function on a pointed CPO $D$. Then $F$ has a least fixpoint $\fix F \definedas \bigsqcup_n F^n(\bot)$. The proof is a direct generalization of the proof for set operators given in Lecture 7, where $\bot$ was $\varnothing$ and $\bigsqcup$ was $\bigcup$. In a nutshell: by monotonicity, the $F^n(\bot)$ form a chain; since $D$ is a CPO, the supremum $\fix F$ of this chain exists; and by continuity, $\fix F$ is preserved by $F$. \section{Flat Domains} Let $S$ be a set with the _discrete ordering_, which means that any two distinct elements of $S$ are $\dlt$-incomparable. We can make $S$ into a pointed CPO $S_\bot$ by adding a new bottom element $\bot$ and defining $\bot\dlt\bot\dlt x\dlt x$ for all $x\in S$, but nothing else. This is called a _flat domain_. For example, $\mathbb N_\bot$ looks like \begin{center} \begin{picture}(0,24)(0,-5) \put(0,-2){\makebox(0,0)[t]{$\bot$}} \put(-5,0){\line(-5,2){26}} \put(-3,0){\line(-5,3){17}} \put(-1,0){\line(-1,1){10}} \put(0,0){\line(0,1){10}} \put(1,0){\line(1,1){10}} \put(3,0){\line(5,3){17}} \put(5,0){\line(5,2){26}} \put(-30,12){\makebox(0,0)[b]{$0$}} \put(-20,12){\makebox(0,0)[b]{$1$}} \put(-10,12){\makebox(0,0)[b]{$2$}} \put(0,12){\makebox(0,0)[b]{$3$}} \put(10,12){\makebox(0,0)[b]{$4$}} \put(20,12){\makebox(0,0)[b]{$5$}} \put(30,12){\makebox(0,0)[b]{$6$}} \put(45,12){\makebox(0,0)[b]{$\cdots$}} \end{picture} \end{center} Any flat domain is chain-complete, since every chain is finite, and every finite nonempty chain has a maximum element, which is its supremum. \section{Continuous Functions on CPOs Form a CPO} Now we claim that if $C$ and $D$ are CPOs, then the space of continuous functions $f:C\to D$ is a CPO under the pointwise ordering \( f \dlt g &\stackrel\triangle\Longleftrightarrow& \forall x\in C\ f(x)\dlt g(x). \) This space is denoted $\cf{C\to D}$. It is easily verified that $\dlt$ is a partial order on $\cf{C\to D}$. If $D$ is pointed with bottom element $\bot$, then $\cf{C\to D}$ is also pointed with bottom element $\bot \definedas \lam{x\in C}\bot$. We need to show that $\cf{C\to D}$ is chain-complete. Let $\mathcal C$ be a nonempty chain in $\cf{C\to D}$. Define \( G &\definedas& \lam{x\in C}{\bigsqcup_{g\in\mathcal C} g(x)}. \) First, $G$ is a well-defined function, since for any $x\in C$, $\{g(x)\mid g\in\mathcal C\}$ is a chain in $D$, therefore its supremum $\bigsqcup_{g\in\mathcal C} g(x)$ exists. Also, the function $G$ is continuous, since for any nonempty chain $E$ in $C$, \reasoning{\bigsqcup_{g\in\mathcal C} g(\bigsqcup E)} \( G(\bigsqcup E) &=& \reason{\bigsqcup_{g\in\mathcal C} g(\bigsqcup E)}{by the definition of $G$}\\ &=& \reason{\bigsqcup_{g\in\mathcal C} \bigsqcup_{x\in E} g(x)}{since each $g\in\mathcal C$ is continuous}\\ &=& \reason{\bigsqcup_{x\in E} \bigsqcup_{g\in\mathcal C} g(x)}{by the lemma below}\\ &=& \reason{\bigsqcup_{x\in E} G(x)}{again by the definition of $G$.} \) The third step in the above argument uses the following lemma. \medskip\noindent \textbf{Lemma} If $a_{xy}$ is a doubly-indexed collection of members of a partially ordered set such that \begin{enumerate} \renewcommand\labelenumi{(\roman{enumi})} \item for all $x$, $\bigsqcup_y a_{xy}$ exists, \item for all $y$, $\bigsqcup_x a_{xy}$ exists, and \item $\bigsqcup_y \bigsqcup_x a_{xy}$ exists, \end{enumerate} then $\bigsqcup_x \bigsqcup_y a_{xy}$ exists and is equal to $\bigsqcup_y \bigsqcup_x a_{xy}$. \medskip _Proof_. Clearly $\bigsqcup_y \bigsqcup_x a_{xy}$ is an upper bound for all $a_{xy}$, therefore it is an upper bound for all $\bigsqcup_y a_{xy}$; and if $b$ is any other upper bound for all $\bigsqcup_y a_{xy}$, then $a_{xy}\dlt b$ for all $x,y$, therefore $\bigsqcup_y \bigsqcup_x a_{xy}\dlt b$, so $\bigsqcup_y \bigsqcup_x a_{xy}$ is the least upper bound for all $\bigsqcup_y a_{xy}$; that is, $\bigsqcup_x \bigsqcup_y a_{xy}=\bigsqcup_y \bigsqcup_x a_{xy}$.\quad$\Box$ \medskip To apply this lemma, we need to know that \begin{enumerate} \renewcommand\labelenumi{(\roman{enumi})} \item for all $g\in\mathcal C$, $\bigsqcup_{x\in E} g(x)$ exists, \item for all $x\in E$, $\bigsqcup_{g\in\mathcal C} g(x)$ exists, and \item $\bigsqcup_{g\in\mathcal C} \bigsqcup_{x\in E} g(x)$ exists. \end{enumerate} But (i) holds because all $g\in\mathcal C$ are continuous, therefore $\bigsqcup_{x\in E} g(x)=g(\bigsqcup\,E)$; (ii) holds because $\{g(x)\mid g\in\mathcal C\}$ is a chain in $D$, and $D$ is chain-complete; and (iii) follows from (i) and (ii) by taking $x=\bigsqcup\,E$. \section{Fixpoints and the Semantics of \textsf{while-do}} Now let's return to the denotational semantics of the "while" loop. We previously defined the function \( F &:& (\Sigma\to\Sigma_\bot)\ \ \to\ \ (\Sigma\to\Sigma_\bot)\\ F &\definedas& \lam{w\in\Sigma\to\Sigma_\bot}{\lam{\sigma\in\Sigma}{\ifthenelse{\Tr Bb\sigma}{(w)^*(\Tr Cc\sigma)}\sigma}}. \) Any function $\Sigma\to\Sigma_\bot$ is continuous, since chains in the discrete space $\Sigma$ contain at most one element, thus the space of functions $\Sigma\to\Sigma_\bot$ is the same as the space of continuous functions $\cf{\Sigma\to\Sigma_\bot}$. Moreover, the lift $(w)^*:\Sigma_\bot\to\Sigma_\bot$ of any function $w:\Sigma\to\Sigma_\bot$ is continuous. By previous arguments, the function space $\cf{\Sigma\to\Sigma_\bot}$ is a pointed CPO, and $F$ maps this space to itself. To obtain a least fixpoint by Knaster--Tarski, we need to know that $F$ is continuous. Let's first check that it is monotone. This will ensure that, when trying to check the definition of continuity, when $C$ is a chain, $\{F(d)\mid d\in C\}$ is also a chain, so that $\bigsqcup_{d\in C} F(d)$ exists. Suppose $d\dlt d'$. We want to show that $F(d)\dlt F(d')$. But for all $\sigma$, \( F(d)(\sigma) &=& \ifthenelse{\Tr Bb\sigma}{(d)^*(\Tr Cc\sigma)}\sigma\\ &\dlt& \ifthenelse{\Tr Bb\sigma}{(d')^*(\Tr Cc\sigma)}\sigma\\ &=& F(d')(\sigma). \) Here we have used the fact that the operator $(\cdot)^*$ is monotone, which is easy to check. Now let's check that $F$ is continuous. Let $C$ be an arbitrary chain. We want to show that $\bigsqcup_{d\in C} F(d) = F(\bigsqcup C)$. We have \( \bigsqcup_{d\in C} F(d) &=& \bigsqcup_{d\in C}\lam{\sigma}{\ifthenelse{\Tr Bb\sigma}{(d)^*(\Tr Cc\sigma)}\sigma}\\ &=& \lam{\sigma}{\bigsqcup_{d\in C}\ifthenelse{\Tr Bb\sigma}{(d)^*(\Tr Cc\sigma)}\sigma}\\ &=& \lam{\sigma}{\ifthenelse{\Tr Bb\sigma}{\bigsqcup_{d\in C} (d)^*(\Tr Cc\sigma)}\sigma}\\ &=& \lam{\sigma}{\ifthenelse{\Tr Bb\sigma}{(\bigsqcup\,C)^*(\Tr Cc\sigma)}\sigma}\ \ =\ \ F(\bigsqcup\,C), \) since $\Tr Bb\sigma$ does not depend on $d$ and since the lift operator $(\cdot)^*$ is continuous. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document}