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Interpreter chapter: expressions and substitution
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frap.tex
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\documentclass{amsbook}
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\usepackage{hyperref,url,amsmath,proof}
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\usepackage{hyperref,url,amsmath,proof,stmaryrd,tikz-cd}
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\newtheorem{theorem}{Theorem}[chapter]
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\newtheorem{lemma}[theorem]{Lemma}
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@ -463,9 +463,100 @@ The general patterns should soon become clear, as they are somehow already famil
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\end{quote}
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The quoted remark could just as well be in Spanish instead of English, in which case we have two languages nested in a nontrivial way.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\chapter{Semantics via Interpreters}
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That's enough about what programs \emph{look like}.
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Let's shift our attention to what programs \emph{mean}.
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\section{Semantics for Arithmetic Expressions via Finite Maps}
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\newcommand{\mempty}[0]{\bullet}
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\newcommand{\msel}[2]{#1(#2)}
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\newcommand{\mupd}[3]{#1[#2 \mapsto #3]}
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To explain the meaning of one of last chapter's arithmetic expressions, we need a way to indicate the value of each variable.
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A theory of \emph{finite maps}\index{finite map} is helpful here.
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We apply the following notations throughout the book: \\
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\begin{tabular}{rl}
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$\mempty$ & empty map, with $\emptyset$ as its domain \\
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$\msel{m}{k}$ & mapping of key $k$ in map $m$ \\
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$\mupd{m}{k}{v}$ & extension of map $m$ to also map key $k$ to value $v$
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\end{tabular} \\
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As the name advertises, finite maps are functions with finite domains, where the domain may be expanded by each extension operation.
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Two axioms explain the essential interactions of the basic operators.
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$$\infer{\msel{\mupd{m}{k}{v}}{k} = v}{}
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\quad
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\infer{\msel{\mupd{m}{k_1}{v}}{k_2} = m(k_2)}{
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k_1 \neq k_2
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}$$
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\newcommand{\denote}[1]{{\left \llbracket #1 \right \rrbracket}}
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With these operators in hand, we can write a semantics for arithmetic expressions.
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This is a recursive function that \emph{maps variable valuations to numbers}.
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We write $\denote{e}$ for the meaning of $e$; this notation is often referred to as \emph{Oxford brackets}\index{Oxford brackets}.
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Recall that we allow notations like this as syntactic sugar for arbitrary functions, even when giving the equations that define those functions.
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We write $v$ for a valuation (finite map).
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\begin{eqnarray*}
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\denote{n}v &=& n \\
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\denote{x}v &=& v(x) \\
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\denote{e_1 + e_2}v &=& \denote{e_1}v + \denote{e_2}v \\
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\denote{e_1 - e_2}v &=& \denote{e_1}v - \denote{e_2}v \\
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\denote{e_1 \times e_2}v &=& \denote{e_1}v \times \denote{e_2}v
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\end{eqnarray*}
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Note how parts of the definition feel a little bit like cheating, as we just ``push notations inside the brackets.''
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It's important to remember that plus \emph{inside} the brackets is syntax, while plus \emph{outside} the brackets is the normal addition of math!
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\newcommand{\subst}[3]{[#3/#2]#1}
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To test our semantics, we define a \emph{variable substitution} function\index{substitution}.
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A substitution $\subst{e}{x}{e'}$ stands for the result of running through the syntax of $e$, replacing every occurrence of variable $x$ with expression $e'$.
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\begin{eqnarray*}
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\subst{n}{x}{e} &=& n \\
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\subst{x}{x}{e} &=& e \\
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\subst{y}{x}{e} &=& y \textrm{, when $y \neq x$} \\
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\subst{(e_1 + e_2)}{x}{e} &=& \subst{e_1}{x}{e} + \subst{e_2}{x}{e} \\
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\subst{(e_1 - e_2)}{x}{e} &=& \subst{e_1}{x}{e} - \subst{e_2}{x}{e} \\
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\subst{(e_1 \times e_2)}{x}{e} &=& \subst{e_1}{x}{e} \times \subst{e_2}{x}{e}
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\end{eqnarray*}
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We can prove a key compatibility property of these two recursive functions.
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\begin{theorem}
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For all $e$, $e'$, $x$, and $v$, $\denote{\subst{e}{x}{e'}}{v} = \denote{e}{(\mupd{v}{x}{\denote{e'}{v}})}$.
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\end{theorem}
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That is, in some sense, the operations of interpretation and substitution \emph{commute} with each other.
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That intuition gives rise to the common notion of a \emph{commuting diagram}\index{commuting diagram}, like the one below for this particular example.
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\[
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\begin{tikzcd}
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(e, v) \arrow{r}{\subst{\ldots}{x}{e'}} \arrow{d}{\mupd{\ldots}{x}{\denote{e'}v}} & (\subst{e}{x}{e'}, v) \arrow{d}{\denote{\ldots}} \\
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(e, \mupd{v}{x}{\denote{e'}v}) \arrow{r}{\denote{\ldots}} & \denote{\subst{e}{x}{e'}}v
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\end{tikzcd}
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\]
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We start at the top left, with a given expresson $e$ and valuation $v$.
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The diagram shows the equivalence of \emph{two different paths} to the bottom right.
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Each individual arrow is labeled with some description of the transformation it performs, to get from the term at its source to the term at its destination.
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The right-then-down path is based on substituting and then interpreting, while the down-then-right path is based on extending the valuation and then interpreting.
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Since both paths wind up at the same spot, the diagram indicates an equality between the corresponding terms.
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It's a matter of taste whether the theorem statement or the diagram expresses the property more clearly!
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\appendix
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\chapter{The Coq Proof Assistant}
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Coq\index{Coq} is a proof-assistant software package developed as open source, primarily by Inria\index{Inria}, the French national computer-science lab.
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