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\begin{isabellebody}%
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\def\isabellecontext{termination}%
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\isamarkupfalse%
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%
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\begin{isamarkuptext}%
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When a function~$f$ is defined via \isacommand{recdef}, Isabelle tries to prove
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its termination with the help of the user-supplied measure. Each of the examples
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above is simple enough that Isabelle can automatically prove that the
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argument's measure decreases in each recursive call. As a result,
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$f$\isa{{\isachardot}simps} will contain the defining equations (or variants derived
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from them) as theorems. For example, look (via \isacommand{thm}) at
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\isa{sep{\isachardot}simps} and \isa{sep{\isadigit{1}}{\isachardot}simps} to see that they define
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the same function. What is more, those equations are automatically declared as
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simplification rules.
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Isabelle may fail to prove the termination condition for some
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recursive call. Let us try to define Quicksort:%
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\end{isamarkuptext}%
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\isamarkuptrue%
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\isacommand{consts}\ qs\ {\isacharcolon}{\isacharcolon}\ {\isachardoublequote}nat\ list\ {\isasymRightarrow}\ nat\ list{\isachardoublequote}\isanewline
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\isamarkupfalse%
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\isacommand{recdef}\ qs\ {\isachardoublequote}measure\ length{\isachardoublequote}\isanewline
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\ {\isachardoublequote}qs\ {\isacharbrackleft}{\isacharbrackright}\ {\isacharequal}\ {\isacharbrackleft}{\isacharbrackright}{\isachardoublequote}\isanewline
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\ {\isachardoublequote}qs{\isacharparenleft}x{\isacharhash}xs{\isacharparenright}\ {\isacharequal}\ qs{\isacharparenleft}filter\ {\isacharparenleft}{\isasymlambda}y{\isachardot}\ y{\isasymle}x{\isacharparenright}\ xs{\isacharparenright}\ {\isacharat}\ {\isacharbrackleft}x{\isacharbrackright}\ {\isacharat}\ qs{\isacharparenleft}filter\ {\isacharparenleft}{\isasymlambda}y{\isachardot}\ x{\isacharless}y{\isacharparenright}\ xs{\isacharparenright}{\isachardoublequote}\isamarkupfalse%
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%
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\begin{isamarkuptext}%
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\noindent where \isa{filter} is predefined and \isa{filter\ P\ xs}
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is the list of elements of \isa{xs} satisfying \isa{P}.
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This definition of \isa{qs} fails, and Isabelle prints an error message
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showing you what it was unable to prove:
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\begin{isabelle}%
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\ \ \ \ \ length\ {\isacharparenleft}filter\ {\isachardot}{\isachardot}{\isachardot}\ xs{\isacharparenright}\ {\isacharless}\ Suc\ {\isacharparenleft}length\ xs{\isacharparenright}%
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\end{isabelle}
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We can either prove this as a separate lemma, or try to figure out which
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existing lemmas may help. We opt for the second alternative. The theory of
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lists contains the simplification rule \isa{length\ {\isacharparenleft}filter\ P\ xs{\isacharparenright}\ {\isasymle}\ length\ xs},
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which is what we need, provided we turn \mbox{\isa{{\isacharless}\ Suc}}
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into
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\isa{{\isasymle}} so that the rule applies. Lemma
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\isa{less{\isacharunderscore}Suc{\isacharunderscore}eq{\isacharunderscore}le} does just that: \isa{{\isacharparenleft}m\ {\isacharless}\ Suc\ n{\isacharparenright}\ {\isacharequal}\ {\isacharparenleft}m\ {\isasymle}\ n{\isacharparenright}}.
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Now we retry the above definition but supply the lemma(s) just found (or
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proved). Because \isacommand{recdef}'s termination prover involves
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simplification, we include in our second attempt a hint: the
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\attrdx{recdef_simp} attribute says to use \isa{less{\isacharunderscore}Suc{\isacharunderscore}eq{\isacharunderscore}le} as a
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simplification rule.\cmmdx{hints}%
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\end{isamarkuptext}%
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\isamarkuptrue%
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\isamarkupfalse%
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\isamarkupfalse%
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\isacommand{recdef}\ qs\ {\isachardoublequote}measure\ length{\isachardoublequote}\isanewline
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\ {\isachardoublequote}qs\ {\isacharbrackleft}{\isacharbrackright}\ {\isacharequal}\ {\isacharbrackleft}{\isacharbrackright}{\isachardoublequote}\isanewline
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\ {\isachardoublequote}qs{\isacharparenleft}x{\isacharhash}xs{\isacharparenright}\ {\isacharequal}\ qs{\isacharparenleft}filter\ {\isacharparenleft}{\isasymlambda}y{\isachardot}\ y{\isasymle}x{\isacharparenright}\ xs{\isacharparenright}\ {\isacharat}\ {\isacharbrackleft}x{\isacharbrackright}\ {\isacharat}\ qs{\isacharparenleft}filter\ {\isacharparenleft}{\isasymlambda}y{\isachardot}\ x{\isacharless}y{\isacharparenright}\ xs{\isacharparenright}{\isachardoublequote}\isanewline
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{\isacharparenleft}\isakeyword{hints}\ recdef{\isacharunderscore}simp{\isacharcolon}\ less{\isacharunderscore}Suc{\isacharunderscore}eq{\isacharunderscore}le{\isacharparenright}\isamarkupfalse%
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\isamarkupfalse%
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%
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\begin{isamarkuptext}%
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\noindent
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This time everything works fine. Now \isa{qs{\isachardot}simps} contains precisely
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the stated recursion equations for \isa{qs} and they have become
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simplification rules.
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Thus we can automatically prove results such as this one:%
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\end{isamarkuptext}%
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\isamarkuptrue%
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\isacommand{theorem}\ {\isachardoublequote}qs{\isacharbrackleft}{\isadigit{2}}{\isacharcomma}{\isadigit{3}}{\isacharcomma}{\isadigit{0}}{\isacharbrackright}\ {\isacharequal}\ qs{\isacharbrackleft}{\isadigit{3}}{\isacharcomma}{\isadigit{0}}{\isacharcomma}{\isadigit{2}}{\isacharbrackright}{\isachardoublequote}\isanewline
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\isamarkupfalse%
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\isamarkupfalse%
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\isamarkupfalse%
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%
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\begin{isamarkuptext}%
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\noindent
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More exciting theorems require induction, which is discussed below.
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If the termination proof requires a lemma that is of general use, you can
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turn it permanently into a simplification rule, in which case the above
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\isacommand{hint} is not necessary. But in the case of
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\isa{less{\isacharunderscore}Suc{\isacharunderscore}eq{\isacharunderscore}le} this would be of dubious value.%
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\end{isamarkuptext}%
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\isamarkuptrue%
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\isamarkupfalse%
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\end{isabellebody}%
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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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