%
\begin{isabellebody}%
\def\isabellecontext{Even}%
\isanewline
\isacommand{theory}\ Even\ {\isacharequal}\ Main{\isacharcolon}\isanewline
\isanewline
\isanewline
\isamarkupfalse%
\isacommand{consts}\ even\ {\isacharcolon}{\isacharcolon}\ {\isachardoublequote}nat\ set{\isachardoublequote}\isanewline
\isamarkupfalse%
\isacommand{inductive}\ even\isanewline
\isakeyword{intros}\isanewline
zero{\isacharbrackleft}intro{\isacharbang}{\isacharbrackright}{\isacharcolon}\ {\isachardoublequote}{\isadigit{0}}\ {\isasymin}\ even{\isachardoublequote}\isanewline
step{\isacharbrackleft}intro{\isacharbang}{\isacharbrackright}{\isacharcolon}\ {\isachardoublequote}n\ {\isasymin}\ even\ {\isasymLongrightarrow}\ {\isacharparenleft}Suc\ {\isacharparenleft}Suc\ n{\isacharparenright}{\isacharparenright}\ {\isasymin}\ even{\isachardoublequote}\isamarkupfalse%
%
\begin{isamarkuptext}%
An inductive definition consists of introduction rules.
\begin{isabelle}%
\ \ \ \ \ n\ {\isasymin}\ Even{\isachardot}even\ {\isasymLongrightarrow}\ Suc\ {\isacharparenleft}Suc\ n{\isacharparenright}\ {\isasymin}\ Even{\isachardot}even%
\end{isabelle}
\rulename{even.step}
\begin{isabelle}%
\ \ \ \ \ {\isasymlbrakk}xa\ {\isasymin}\ Even{\isachardot}even{\isacharsemicolon}\ P\ {\isadigit{0}}{\isacharsemicolon}\ {\isasymAnd}n{\isachardot}\ {\isasymlbrakk}n\ {\isasymin}\ Even{\isachardot}even{\isacharsemicolon}\ P\ n{\isasymrbrakk}\ {\isasymLongrightarrow}\ P\ {\isacharparenleft}Suc\ {\isacharparenleft}Suc\ n{\isacharparenright}{\isacharparenright}{\isasymrbrakk}\isanewline
\isaindent{\ \ \ \ \ }{\isasymLongrightarrow}\ P\ xa%
\end{isabelle}
\rulename{even.induct}
Attributes can be given to the introduction rules. Here both rules are
specified as \isa{intro!}
Our first lemma states that numbers of the form $2\times k$ are even.%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ two{\isacharunderscore}times{\isacharunderscore}even{\isacharbrackleft}intro{\isacharbang}{\isacharbrackright}{\isacharcolon}\ {\isachardoublequote}{\isadigit{2}}{\isacharasterisk}k\ {\isasymin}\ even{\isachardoublequote}\isanewline
\isamarkupfalse%
\isamarkupfalse%
\isamarkuptrue%
\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
Our goal is to prove the equivalence between the traditional definition
of even (using the divides relation) and our inductive definition. Half of
this equivalence is trivial using the lemma just proved, whose \isa{intro!}
attribute ensures it will be applied automatically.%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ dvd{\isacharunderscore}imp{\isacharunderscore}even{\isacharcolon}\ {\isachardoublequote}{\isadigit{2}}\ dvd\ n\ {\isasymLongrightarrow}\ n\ {\isasymin}\ even{\isachardoublequote}\isanewline
\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
our first rule induction!%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ even{\isacharunderscore}imp{\isacharunderscore}dvd{\isacharcolon}\ {\isachardoublequote}n\ {\isasymin}\ even\ {\isasymLongrightarrow}\ {\isadigit{2}}\ dvd\ n{\isachardoublequote}\isanewline
\isamarkupfalse%
\isamarkupfalse%
\isamarkuptrue%
\isamarkupfalse%
\isamarkuptrue%
\isamarkupfalse%
\isamarkuptrue%
\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
no iff-attribute because we don't always want to use it%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{theorem}\ even{\isacharunderscore}iff{\isacharunderscore}dvd{\isacharcolon}\ {\isachardoublequote}{\isacharparenleft}n\ {\isasymin}\ even{\isacharparenright}\ {\isacharequal}\ {\isacharparenleft}{\isadigit{2}}\ dvd\ n{\isacharparenright}{\isachardoublequote}\isanewline
\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
this result ISN'T inductive...%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ Suc{\isacharunderscore}Suc{\isacharunderscore}even{\isacharunderscore}imp{\isacharunderscore}even{\isacharcolon}\ {\isachardoublequote}Suc\ {\isacharparenleft}Suc\ n{\isacharparenright}\ {\isasymin}\ even\ {\isasymLongrightarrow}\ n\ {\isasymin}\ even{\isachardoublequote}\isanewline
\isamarkupfalse%
\isamarkupfalse%
\isamarkuptrue%
\isamarkupfalse%
%
\begin{isamarkuptext}%
...so we need an inductive lemma...%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ even{\isacharunderscore}imp{\isacharunderscore}even{\isacharunderscore}minus{\isacharunderscore}{\isadigit{2}}{\isacharcolon}\ {\isachardoublequote}n\ {\isasymin}\ even\ {\isasymLongrightarrow}\ n\ {\isacharminus}\ {\isadigit{2}}\ {\isasymin}\ even{\isachardoublequote}\isanewline
\isamarkupfalse%
\isamarkupfalse%
\isamarkuptrue%
\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
...and prove it in a separate step%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ Suc{\isacharunderscore}Suc{\isacharunderscore}even{\isacharunderscore}imp{\isacharunderscore}even{\isacharcolon}\ {\isachardoublequote}Suc\ {\isacharparenleft}Suc\ n{\isacharparenright}\ {\isasymin}\ even\ {\isasymLongrightarrow}\ n\ {\isasymin}\ even{\isachardoublequote}\isanewline
\isamarkupfalse%
\isanewline
\isanewline
\isamarkupfalse%
\isacommand{lemma}\ {\isacharbrackleft}iff{\isacharbrackright}{\isacharcolon}\ {\isachardoublequote}{\isacharparenleft}{\isacharparenleft}Suc\ {\isacharparenleft}Suc\ n{\isacharparenright}{\isacharparenright}\ {\isasymin}\ even{\isacharparenright}\ {\isacharequal}\ {\isacharparenleft}n\ {\isasymin}\ even{\isacharparenright}{\isachardoublequote}\isanewline
\isamarkupfalse%
\isanewline
\isamarkupfalse%
\isacommand{end}\isanewline
\isanewline
\isamarkupfalse%
\end{isabellebody}%
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