(*<*)

theory case_exprs = Main:

(*>*)

subsection{*Case expressions*}

text{*\label{sec:case-expressions}

HOL also features \isaindexbold{case}-expressions for analyzing

elements of a datatype. For example,

@{term[display]"case xs of [] => 1 | y#ys => y"}

evaluates to @{term"1"} if @{term"xs"} is @{term"[]"} and to @{term"y"} if 

@{term"xs"} is @{term"y#ys"}. (Since the result in both branches must be of

the same type, it follows that @{term"y"} is of type @{typ"nat"} and hence

that @{term"xs"} is of type @{typ"nat list"}.)

In general, if $e$ is a term of the datatype $t$ defined in

\S\ref{sec:general-datatype} above, the corresponding

@{text"case"}-expression analyzing $e$ is

\[

\begin{array}{rrcl}

@{text"case"}~e~@{text"of"} & C@1~x@ {11}~\dots~x@ {1k@1} & \To & e@1 \\

                           \vdots \\

                           \mid & C@m~x@ {m1}~\dots~x@ {mk@m} & \To & e@m

\end{array}

\]

\begin{warn}

\emph{All} constructors must be present, their order is fixed, and nested

patterns are not supported.  Violating these restrictions results in strange

error messages.

\end{warn}

\noindent

Nested patterns can be simulated by nested @{text"case"}-expressions: instead

of

@{text[display]"case xs of [] => 1 | [x] => x | x # (y # zs) => y"}

write

@{term[display,eta_contract=false,margin=50]"case xs of [] => 1 | x#ys => (case ys of [] => x | y#zs => y)"}

Note that @{text"case"}-expressions may need to be enclosed in parentheses to

indicate their scope

*}

subsection{*Structural induction and case distinction*}

text{*

\indexbold{structural induction}

\indexbold{induction!structural}

\indexbold{case distinction}

Almost all the basic laws about a datatype are applied automatically during

simplification. Only induction is invoked by hand via \isaindex{induct_tac},

which works for any datatype. In some cases, induction is overkill and a case

distinction over all constructors of the datatype suffices. This is performed

by \isaindexbold{case_tac}. A trivial example:

*}

lemma "(case xs of [] \<Rightarrow> [] | y#ys \<Rightarrow> xs) = xs";

apply(case_tac xs);

txt{*\noindent

results in the proof state

\begin{isabelle}

~1.~xs~=~[]~{\isasymLongrightarrow}~(case~xs~of~[]~{\isasymRightarrow}~[]~|~y~\#~ys~{\isasymRightarrow}~xs)~=~xs\isanewline

~2.~{\isasymAnd}a~list.~xs=a\#list~{\isasymLongrightarrow}~(case~xs~of~[]~{\isasymRightarrow}~[]~|~y\#ys~{\isasymRightarrow}~xs)~=~xs%

\end{isabelle}

which is solved automatically:

*}

apply(auto)

(*<*)done(*>*)

text{*

Note that we do not need to give a lemma a name if we do not intend to refer

to it explicitly in the future.

*}

(*<*)

end

(*>*)