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(*<*)
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theory CodeGen = Main:
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(*>*)
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text{*\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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*}
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types 'v binop = "'v \\<Rightarrow> 'v \\<Rightarrow> 'v";
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datatype ('a,'v)expr = Cex 'v
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| Vex 'a
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| Bex "'v binop" "('a,'v)expr" "('a,'v)expr";
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text{*\noindent
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The three constructors represent constants, variables and the combination of
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two subexpressions with a binary operation.
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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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*}
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consts value :: "('a \\<Rightarrow> 'v) \\<Rightarrow> ('a,'v)expr \\<Rightarrow> 'v";
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primrec
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"value env (Cex v) = v"
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"value env (Vex a) = env a"
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"value env (Bex f e1 e2) = f (value env e1) (value env e2)";
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text{*
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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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*}
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datatype ('a,'v) instr = Const 'v
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| Load 'a
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| Apply "'v binop";
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text{*
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The execution of the stack machine is modelled by a function \isa{exec}
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that takes a store (modelled as a function from addresses to values, just
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like the environment for evaluating expressions), a stack (modelled as a
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list) of values, and a list of instructions, and returns the stack at the end
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of the execution---the store remains unchanged:
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*}
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consts exec :: "('a\\<Rightarrow>'v) \\<Rightarrow> 'v list \\<Rightarrow> ('a,'v)instr list \\<Rightarrow> 'v list";
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primrec
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"exec s vs [] = vs"
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"exec s vs (i#is) = (case i of
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Const v \\<Rightarrow> exec s (v#vs) is
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| Load a \\<Rightarrow> exec s ((s a)#vs) is
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| Apply f \\<Rightarrow> exec s ( (f (hd vs) (hd(tl vs)))#(tl(tl vs)) ) is)";
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text{*\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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*}
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consts comp :: "('a,'v)expr \\<Rightarrow> ('a,'v)instr list";
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primrec
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"comp (Cex v) = [Const v]"
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"comp (Vex a) = [Load a]"
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"comp (Bex f e1 e2) = (comp e2) @ (comp e1) @ [Apply f]";
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text{*
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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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*}
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theorem "exec s [] (comp e) = [value s e]";
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(*<*)oops;(*>*)
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text{*\noindent
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This theorem needs to be generalized to
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*}
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theorem "\\<forall>vs. exec s vs (comp e) = (value s e) # vs";
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txt{*\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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*}
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(*<*)oops;(*>*)
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lemma exec_app[simp]:
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"\\<forall>vs. exec s vs (xs@ys) = exec s (exec s vs xs) ys";
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txt{*\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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*}
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apply(induct_tac xs, simp, simp split: instr.split).;
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text{*\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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*}
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(*<*)
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lemmas [simp del] = exec_app;
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lemma [simp]: "\\<forall>vs. exec s vs (xs@ys) = exec s (exec s vs xs) ys";
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(*>*)
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apply(induct_tac xs, auto split: instr.split).;
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text{*\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 s [] (comp e) = [value s e]}
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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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*}
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(*<*)
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theorem "\\<forall>vs. exec s vs (comp e) = (value s e) # vs";
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apply(induct_tac e, auto).;
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end
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(*>*)
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