doc-src/Exercises/2002/a1/generated/a1.tex
author kleing
Sat, 01 Mar 2003 19:34:54 +0100
changeset 13844 44f741cdcea3
parent 13841 ed4e97874454
permissions -rw-r--r--
typo

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\begin{isabellebody}%
\def\isabellecontext{a{\isadigit{1}}}%
\isamarkupfalse%
%
\isamarkupsubsection{Lists%
}
\isamarkuptrue%
%
\begin{isamarkuptext}%
Define a universal and an existential quantifier on lists.
Expression \isa{alls\ P\ xs} should be true iff \isa{P\ x} holds
for every element \isa{x} of \isa{xs}, and \isa{exs\ P\ xs}
should be true iff \isa{P\ x} holds for some element \isa{x} of
\isa{xs}.%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{consts}\ \isanewline
\ \ alls\ {\isacharcolon}{\isacharcolon}\ {\isachardoublequote}{\isacharparenleft}{\isacharprime}a\ {\isasymRightarrow}\ bool{\isacharparenright}\ {\isasymRightarrow}\ {\isacharprime}a\ list\ {\isasymRightarrow}\ bool{\isachardoublequote}\isanewline
\ \ exs\ \ {\isacharcolon}{\isacharcolon}\ {\isachardoublequote}{\isacharparenleft}{\isacharprime}a\ {\isasymRightarrow}\ bool{\isacharparenright}\ {\isasymRightarrow}\ {\isacharprime}a\ list\ {\isasymRightarrow}\ bool{\isachardoublequote}\isamarkupfalse%
%
\begin{isamarkuptext}%
Prove or disprove (by counter example) the following theorems.
You may have to prove some lemmas first.

Use the \isa{{\isacharbrackleft}simp{\isacharbrackright}}-attribute only if the equation is truly a
simplification and is necessary for some later proof.%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ {\isachardoublequote}alls\ {\isacharparenleft}{\isasymlambda}x{\isachardot}\ P\ x\ {\isasymand}\ Q\ x{\isacharparenright}\ xs\ {\isacharequal}\ {\isacharparenleft}alls\ P\ xs\ {\isasymand}\ alls\ Q\ xs{\isacharparenright}{\isachardoublequote}\isamarkupfalse%
\isanewline
\isamarkupfalse%
\isacommand{lemma}\ {\isachardoublequote}alls\ P\ {\isacharparenleft}rev\ xs{\isacharparenright}\ {\isacharequal}\ alls\ P\ xs{\isachardoublequote}\isamarkupfalse%
\isanewline
\isamarkupfalse%
\isacommand{lemma}\ {\isachardoublequote}exs\ {\isacharparenleft}{\isasymlambda}x{\isachardot}\ P\ x\ {\isasymand}\ Q\ x{\isacharparenright}\ xs\ {\isacharequal}\ {\isacharparenleft}exs\ P\ xs\ {\isasymand}\ exs\ Q\ xs{\isacharparenright}{\isachardoublequote}\isamarkupfalse%
\isanewline
\isamarkupfalse%
\isacommand{lemma}\ {\isachardoublequote}exs\ P\ {\isacharparenleft}map\ f\ xs{\isacharparenright}\ {\isacharequal}\ exs\ {\isacharparenleft}P\ o\ f{\isacharparenright}\ xs{\isachardoublequote}\isamarkupfalse%
\isanewline
\isamarkupfalse%
\isacommand{lemma}\ {\isachardoublequote}exs\ P\ {\isacharparenleft}rev\ xs{\isacharparenright}\ {\isacharequal}\ exs\ P\ xs{\isachardoublequote}\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
Find a term \isa{Z} such that the following equation holds:%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ {\isachardoublequote}exs\ {\isacharparenleft}{\isasymlambda}x{\isachardot}\ P\ x\ {\isasymor}\ Q\ x{\isacharparenright}\ xs\ {\isacharequal}\ Z{\isachardoublequote}\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
Express the existential via the universal quantifier ---
\isa{exs} should not occur on the right-hand side:%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ {\isachardoublequote}exs\ P\ xs\ {\isacharequal}\ Z{\isachardoublequote}\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
Define a function \isa{is{\isacharunderscore}in\ x\ xs} that checks if \isa{x} occurs in
\isa{xs}. Now express \isa{is{\isacharunderscore}in} via \isa{exs}:%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ {\isachardoublequote}is{\isacharunderscore}in\ a\ xs\ {\isacharequal}\ Z{\isachardoublequote}\isamarkupfalse%
\isamarkupfalse%
%
\begin{isamarkuptext}%
Define a function \isa{nodups\ xs} that is true iff
\isa{xs} does not contain duplicates, and a function \isa{deldups\ xs} that removes all duplicates. Note that \isa{deldups\ {\isacharbrackleft}x{\isacharcomma}\ y{\isacharcomma}\ x{\isacharbrackright}}
(where \isa{x} and \isa{y} are distinct) can be either
\isa{{\isacharbrackleft}x{\isacharcomma}\ y{\isacharbrackright}} or \isa{{\isacharbrackleft}y{\isacharcomma}\ x{\isacharbrackright}}.

Prove or disprove (by counter example) the following theorems.%
\end{isamarkuptext}%
\isamarkuptrue%
\isacommand{lemma}\ {\isachardoublequote}length\ {\isacharparenleft}deldups\ xs{\isacharparenright}\ {\isacharless}{\isacharequal}\ length\ xs{\isachardoublequote}\isamarkupfalse%
\isanewline
\isamarkupfalse%
\isacommand{lemma}\ {\isachardoublequote}nodups\ {\isacharparenleft}deldups\ xs{\isacharparenright}{\isachardoublequote}\isamarkupfalse%
\isanewline
\isamarkupfalse%
\isacommand{lemma}\ {\isachardoublequote}deldups\ {\isacharparenleft}rev\ xs{\isacharparenright}\ {\isacharequal}\ rev\ {\isacharparenleft}deldups\ xs{\isacharparenright}{\isachardoublequote}\isamarkupfalse%
\isamarkupfalse%
\isamarkupfalse%
\end{isabellebody}%
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