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OperationalSemantics chapter: small-step
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frap_book.tex
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frap_book.tex
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@ -1322,6 +1322,7 @@ $$\begin{array}{rrcl}
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For our example relation, we will define a relation written $\bigstep{(v, c)}{v'}$, for ``command $c$, run with variable valuation $v$, terminates, modifying the valuation to $v'$.''
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This relation is fairly straightforward to define with inference rules.
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\encoding
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$$\infer{\bigstep{(v, \skipe)}{v}}{}
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\quad \infer{\bigstep{(v, \assign{x}{e})}{\mupd{v}{x}{\denote{e}v}}}{}
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\quad \infer{\bigstep{(v, c_1; c_2)}{v_2}}{
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@ -1388,6 +1389,135 @@ Most of our program proofs in this book establish \emph{safety properties}\index
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However, these last two examples with big-step semantics also establish program termination, taking us a few steps into the world of \emph{liveness properties}\index{liveness properties}.
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\section{Small-Step Semantics}
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Often it is convenient to break a system's execution into small sequential steps, rather than executing a whole program in one go.
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Perhaps the most compelling example comes from concurrency, where it is difficult to give a big-step semantics directly.
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Nonterminating programs are the other standard example.
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We want to be able to establish invariants for those programs, all the same, and we need a semantics to help us state what it means to be an invariant.
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\newcommand{\smallstep}[2]{#1 \to #2}
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The canonical solution is \emph{small-step operational semantics}\index{small-step operational semantics}, probably the most common approach to formal program semantics in contemporary research.
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Now we define a single-step relation $\smallstep{(v, c)}{(v', c')}$, meaning that one execution step transforms the first state into the second state.
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Each state is a valuation $v$ and a current command $c$.
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These inference rules give the details.
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\encoding
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$$\infer{\smallstep{(v, \assign{x}{e})}{(\mupd{v}{x}{\denote{e}v}, \skipe)}}{}
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\quad \infer{\smallstep{(v, c_1; c_2)}{(v', c'_1; c_2)}}{
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\smallstep{(v, c_1)}{(v', c'_1)}
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}
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\quad \infer{\smallstep{(v, \skipe; c_2)}{(v, c_2)}}{}$$
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$$\infer{\smallstep{(v, \ifte{e}{c_1}{c_2})}{(v, c_1)}}{
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\denote{e}v \neq 0
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}
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\quad \infer{\smallstep{(v, \ifte{e}{c_1}{c_2})}{(v, c_2)}}{
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\denote{e}v = 0
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}$$
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$$\infer{\smallstep{(v, \while{e}{c_1})}{(v, c_1; \while{e}{c_1})}}{
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\denote{e}v \neq 0
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}
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\quad \infer{\smallstep{(v, \while{e}{c_1})}{(v, \skipe)}}{
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\denote{e}v = 0
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}$$
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The intuition behind the rules may come best from working out an example.
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\newcommand{\smallsteps}[2]{#1 \to^* #2}
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\begin{theorem}
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There exists valuation $v$ such that $\smallsteps{(\mupd{\mempty}{\mathtt{input}}{2}, \mathtt{factorial})}{(v, \skipe)}$ and $\msel{v}{\mathtt{output}} = 2$.
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\end{theorem}
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\begin{proof}
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Here is a step-by-step (literally!) derivation that finds $v$.
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$$\begin{array}{cl}
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& (\mupd{\mempty}{\mathtt{input}}{2}, \assign{\mathtt{output}}{1}; \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{2}}{\mathtt{output}}{1}, \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{2}}{\mathtt{output}}{1}, (\assign{\mathtt{output}}{\mathtt{output} \times \mathtt{input}}; \assign{\mathtt{input}}{\mathtt{input} - 1}); \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{2}}{\mathtt{output}}{2}, (\skipe; \assign{\mathtt{input}}{\mathtt{input} - 1}); \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{2}}{\mathtt{output}}{2}, \assign{\mathtt{input}}{\mathtt{input} - 1}; \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{1}}{\mathtt{output}}{2}, \skipe; \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{1}}{\mathtt{output}}{2}, \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{1}}{\mathtt{output}}{2}, (\assign{\mathtt{output}}{\mathtt{output} \times \mathtt{input}}; \assign{\mathtt{input}}{\mathtt{input} - 1}); \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{1}}{\mathtt{output}}{2}, (\skipe; \assign{\mathtt{input}}{\mathtt{input} - 1}); \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{1}}{\mathtt{output}}{2}, \assign{\mathtt{input}}{\mathtt{input} - 1}; \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{0}}{\mathtt{output}}{2}, \skipe; \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{0}}{\mathtt{output}}{2}, \mathtt{factorial\_loop}) \\
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\to & (\mupd{\mupd{\mempty}{\mathtt{input}}{0}}{\mathtt{output}}{2}, \skipe)
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\end{array}$$
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Clearly the final valuation assigns $\mathtt{output}$ to 2.
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\end{proof}
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\subsection{Equivalence of Big-Step and Small-Step}
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Different theorems are easier to prove with different semantics, so it is helpful to establish formally the intuitive connection between big and small steps.
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\begin{lemma}
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If $\smallsteps{(v, c_1)}{(v', c'_1)}$, then $\smallsteps{(v, c_1; c_2)}{(v', c'_1; c_2)}$,
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\end{lemma}
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\begin{proof}
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By induction on the derivation of $\smallsteps{(v, c_1)}{(v', c'_1)}$.
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\end{proof}
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\begin{theorem}
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If $\bigstep{(v, c)}{v'}$, then $\smallsteps{(v, c)}{(v', \skipe)}$.
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\end{theorem}
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\begin{proof}
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By induction on the derivation of $\bigstep{(v, c)}{v'}$, appealing to the last lemma at two points.
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\end{proof}
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\begin{lemma}
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If $\smallstep{(v, c)}{(v', c')}$ and $\bigstep{(v', c')}{v''}$, then $\bigstep{(v, c)}{v''}$. In other words, we can add a small step to the beginning of any big-step derivation.
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\end{lemma}
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\begin{proof}
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By induction on the derivation of $\smallstep{(v, c)}{(v', c')}$.
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\end{proof}
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\begin{lemma}
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If $\smallsteps{(v, c)}{(v', c')}$ and $\bigstep{(v', c')}{v''}$, then $\bigstep{(v, c)}{v''}$. In other words, we can add any number of small steps to the beginning of any big-step derivation.
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\end{lemma}
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\begin{proof}
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By induction on the derivation of $\smallsteps{(v, c)}{(v', c')}$, appealing to the last lemma.
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\end{proof}
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\begin{theorem}
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If $\smallsteps{(v, c)}{(v', \skipe)}$, then $\bigstep{(v, c)}{v'}$.
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\end{theorem}
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\begin{proof}
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Largely by appeal to the last lemma, considering that $\bigstep{(v', \skipe)}{v'}$.
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\end{proof}
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\subsection{Transition Systems from Small-Step Semantics}
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The small-step semantics is a natural fit with our working definition of transition systems.
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We can define a transition system from any valuation and command, where $\mathbb V$ is the set of valuations and $\mathbb C$ the set of commands, by $\mathbb T(v, c) = \angled{\mathbb V \times \mathbb C, {(v, c)}, \to}$.
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Now we bring to bear all of our machinery about invariants and their proof methods.
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For instance, consider program $P = \while{\mathtt{n}}{\assign{\mathtt{a}}{\mathtt{a} + \mathtt{n}}; \assign{\mathtt{n}}{\mathtt{n} - 2}}$.
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\invariants
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\begin{theorem}
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For $\mathbb T(\mupd{\mupd{\mempty}{\mathtt{n}}{0}}{\mathtt{a}}{0}, P)$, it is an invariant that the valuation maps variable $\mathtt{a}$ to an even number.
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\end{theorem}
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\begin{proof}
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First, we strengthen the invariant.
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We compute the set $\overline{P}$ of all commands that can be reached from $P$ by stepping the small-step semantics.
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This set is finite, even though the set of \emph{reachable valuations} is infinite.
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Our strengthened invariant is $I(v, c) = c \in \overline{P} \land (\exists n. \; \msel{v}{\mathtt{n}} = n \land \textrm{even}(n)) \land (\exists a. \; \msel{v}{\mathtt{a}} = a \land \textrm{even}(a))$.
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In other words, we strengthen by adding the constraints that (1) we do not stray from the expected set of reachable commands and (2) variable \texttt{n} also remains even.
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The strengthened invariant is straightforward to prove by invariant induction, using repeated inversion on $\to$ facts.
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\end{proof}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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