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\begin{isabelle}%
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%
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\begin{isamarkuptext}%
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\noindent
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The task is to develop a compiler from a generic type of expressions (built
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up from variables, constants and binary operations) to a stack machine. This
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generic type of expressions is a generalization of the boolean expressions in
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\S\ref{sec:boolex}. This time we do not commit ourselves to a particular
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type of variables or values but make them type parameters. Neither is there
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a fixed set of binary operations: instead the expression contains the
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appropriate function itself.%
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\end{isamarkuptext}%
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\isacommand{types}~'v~binop~=~{"}'v~{\isasymRightarrow}~'v~{\isasymRightarrow}~'v{"}\isanewline
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\isacommand{datatype}~('a,'v)expr~=~Cex~'v\isanewline
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~~~~~~~~~~~~~~~~~~~~~|~Vex~'a\isanewline
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~~~~~~~~~~~~~~~~~~~~~|~Bex~{"}'v~binop{"}~~{"}('a,'v)expr{"}~~{"}('a,'v)expr{"}%
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\begin{isamarkuptext}%
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\noindent
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The three constructors represent constants, variables and the application of
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a binary operation to two subexpressions.
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The value of an expression w.r.t.\ an environment that maps variables to
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values is easily defined:%
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\end{isamarkuptext}%
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\isacommand{consts}~value~::~{"}('a,'v)expr~{\isasymRightarrow}~('a~{\isasymRightarrow}~'v)~{\isasymRightarrow}~'v{"}\isanewline
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\isacommand{primrec}\isanewline
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{"}value~(Cex~v)~env~=~v{"}\isanewline
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{"}value~(Vex~a)~env~=~env~a{"}\isanewline
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{"}value~(Bex~f~e1~e2)~env~=~f~(value~e1~env)~(value~e2~env){"}%
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\begin{isamarkuptext}%
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The stack machine has three instructions: load a constant value onto the
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stack, load the contents of a certain address onto the stack, and apply a
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binary operation to the two topmost elements of the stack, replacing them by
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the result. As for \isa{expr}, addresses and values are type parameters:%
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\end{isamarkuptext}%
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\isacommand{datatype}~('a,'v)~instr~=~Const~'v\isanewline
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~~~~~~~~~~~~~~~~~~~~~~~|~Load~'a\isanewline
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~~~~~~~~~~~~~~~~~~~~~~~|~Apply~{"}'v~binop{"}%
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\begin{isamarkuptext}%
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The execution of the stack machine is modelled by a function
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\isa{exec} that takes a list of instructions, a store (modelled as a
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function from addresses to values, just like the environment for
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evaluating expressions), and a stack (modelled as a list) of values,
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and returns the stack at the end of the execution---the store remains
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unchanged:%
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\end{isamarkuptext}%
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\isacommand{consts}~exec~::~{"}('a,'v)instr~list~{\isasymRightarrow}~('a{\isasymRightarrow}'v)~{\isasymRightarrow}~'v~list~{\isasymRightarrow}~'v~list{"}\isanewline
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\isacommand{primrec}\isanewline
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{"}exec~[]~s~vs~=~vs{"}\isanewline
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{"}exec~(i\#is)~s~vs~=~(case~i~of\isanewline
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~~~~Const~v~~{\isasymRightarrow}~exec~is~s~(v\#vs)\isanewline
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~~|~Load~a~~~{\isasymRightarrow}~exec~is~s~((s~a)\#vs)\isanewline
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~~|~Apply~f~~{\isasymRightarrow}~exec~is~s~((f~(hd~vs)~(hd(tl~vs)))\#(tl(tl~vs)))){"}%
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\begin{isamarkuptext}%
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\noindent
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Recall that \isa{hd} and \isa{tl}
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return the first element and the remainder of a list.
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Because all functions are total, \isa{hd} is defined even for the empty
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list, although we do not know what the result is. Thus our model of the
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machine always terminates properly, although the above definition does not
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tell us much about the result in situations where \isa{Apply} was executed
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with fewer than two elements on the stack.
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The compiler is a function from expressions to a list of instructions. Its
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definition is pretty much obvious:%
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\end{isamarkuptext}%
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\isacommand{consts}~comp~::~{"}('a,'v)expr~{\isasymRightarrow}~('a,'v)instr~list{"}\isanewline
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\isacommand{primrec}\isanewline
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{"}comp~(Cex~v)~~~~~~~=~[Const~v]{"}\isanewline
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{"}comp~(Vex~a)~~~~~~~=~[Load~a]{"}\isanewline
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{"}comp~(Bex~f~e1~e2)~=~(comp~e2)~@~(comp~e1)~@~[Apply~f]{"}%
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\begin{isamarkuptext}%
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Now we have to prove the correctness of the compiler, i.e.\ that the
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execution of a compiled expression results in the value of the expression:%
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\end{isamarkuptext}%
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\isacommand{theorem}~{"}exec~(comp~e)~s~[]~=~[value~e~s]{"}%
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\begin{isamarkuptext}%
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\noindent
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This theorem needs to be generalized to%
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\end{isamarkuptext}%
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\isacommand{theorem}~{"}{\isasymforall}vs.~exec~(comp~e)~s~vs~=~(value~e~s)~\#~vs{"}%
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\begin{isamarkuptxt}%
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\noindent
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which is proved by induction on \isa{e} followed by simplification, once
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we have the following lemma about executing the concatenation of two
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instruction sequences:%
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\end{isamarkuptxt}%
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\isacommand{lemma}~exec\_app[simp]:\isanewline
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~~{"}{\isasymforall}vs.~exec~(xs@ys)~s~vs~=~exec~ys~s~(exec~xs~s~vs){"}%
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\begin{isamarkuptxt}%
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\noindent
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This requires induction on \isa{xs} and ordinary simplification for the
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base cases. In the induction step, simplification leaves us with a formula
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that contains two \isa{case}-expressions over instructions. Thus we add
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automatic case splitting as well, which finishes the proof:%
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\end{isamarkuptxt}%
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\isacommand{by}(induct\_tac~xs,~simp,~simp~split:~instr.split)%
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\begin{isamarkuptext}%
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\noindent
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Note that because \isaindex{auto} performs simplification, it can
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also be modified in the same way \isa{simp} can. Thus the proof can be
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rewritten as%
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\end{isamarkuptext}%
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\isacommand{by}(induct\_tac~xs,~auto~split:~instr.split)%
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\begin{isamarkuptext}%
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\noindent
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Although this is more compact, it is less clear for the reader of the proof.
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We could now go back and prove \isa{exec (comp e) s [] = [value e s]}
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merely by simplification with the generalized version we just proved.
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However, this is unnecessary because the generalized version fully subsumes
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its instance.%
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\end{isamarkuptext}%
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\end{isabelle}%
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: "root"
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%%% End:
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