### bla

parent 2a640499
 ... ... @@ -219,6 +219,8 @@ variant with more compact derivations. \eras {\tcase{e}{\tau}{e_1}{e_2}}&= \tcase{\eras e}{\tau}{\eras{e_1}}{\eras{e_2}}\\ \eras {\letexp{x{:}\anns}{e_1}{e_2}} &= \eras{e_2}\subs{x}{\eras{e_1}} \end{align*} (notice the rule for the let). \subsubsection{Soundness of the algorithm} If $\Gamma\vdashA e:t$ then $\Gamma\vdash \eras e: t$ \subsubsection{Completeness of the algorithm} ... ... @@ -231,7 +233,7 @@ $e\reduces e'$ if and only if $\eras e \reduces \eras{e}$ \subsection{Normalization} To every source term $e$ there exists many algorithmic terms $e'$ such For every source term $e$ there exist many algorithmic terms $e'$ such that $e=\eras{e'}$. The various $e'$ may vary for the annotations used and the use of let's. We want to eliminate the latter possibility of variability and consider only terms in which the let expressions are ... ... @@ -248,7 +250,7 @@ in a particular form defined as follows. \end{array} \end{equation} Now it is possible to associate to every \emph{well-typed} explicitly-typed term $e$ a explicitly-typed term $e$ a particular normal form that has the same type and the same reduction semantics. This is done by induction on the derivation of the type for $e$. Since every well-typed explicitly-typed term has a unique type ... ... @@ -256,16 +258,18 @@ $e$. Since every well-typed explicitly-typed term has a unique type the terms, where $\typof{e}$ denotes the unique type derived for $e$. \begin{align*} \norm {\lambda x{:}\anns.e} & = \lambda x{:}\anns.\norm e\\ \norm {e_1e_2} &= \letexp{x_1{:}\typof{e_1}}{\norm{e_1}}{\letexp{x_2{:}\typof{e_1}}{\norm{e_2}}{x_1x_2}}\\ \norm {(e_1,e_2)} &= \letexp{x_1{:}\typof{e_1}}{\norm{e_1}}{\letexp{x_2{:}\typof{e_1}}{\norm{e_2}}{(x_1,x_2)}}\\ \norm {\pi_i e}&= \letexp{x{:}\typof{e}}{\norm{e}}{\pi_ix} & i=1,2\\ \norm {\tcase{e}{\tau}{e_1}{e_2}}&= \letexp{x{:}\typof{e}}{\norm{e}}{\tcase{x}{\tau}{\norm{e_1}}{\norm{e_2}}}\\ \norm {e_1e_2} &= \letexp{x_1{:}\typof{e_1}}{\norm{e_1}}{\letexp{x_2{:}\typof{e_1}}{\norm{e_2}}{x_1x_2}} &x_1,x_2\text{ fresh}\\ \norm {(e_1,e_2)} &= \letexp{x_1{:}\typof{e_1}}{\norm{e_1}}{\letexp{x_2{:}\typof{e_1}}{\norm{e_2}}{(x_1,x_2)}} &x_1,x_2\text{ fresh}\\ \norm {\pi_i e}&= \letexp{x{:}\typof{e}}{\norm{e}}{\pi_ix} \qquad x\text{ fresh}& i=1,2\\ \norm {\tcase{e}{\tau}{e_1}{e_2}}&= \letexp{x{:}\typof{e}}{\norm{e}}{\tcase{x}{\tau}{\norm{e_1}}{\norm{e_2}}}&x\text{ fresh}\\ \norm {e} &= e &\text{otherwise} \end{align*} The case apply only when the various $e$ $e_i$ are not already The cases apply only when the various $e$ $e_i$ are not already variables (all the other cases are long to write). Notice that $\eras e = \eras{\norm e}$. Notice that $\eras e = \eras{\norm e}$ (and that they have the same type). Therefore what we have is that the soundness but, above all, the completeness of the algorithm can be stated for normal forms: ... ...
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