doc-src/Functions/Thy/Functions.thy

+(*  Title:      doc-src/IsarAdvanced/Functions/Thy/Fundefs.thy
+    Author:     Alexander Krauss, TU Muenchen
+
+Tutorial for function definitions with the new "function" package.
+*)
+
+theory Functions
+imports Main
+begin
+
+section {* Function Definitions for Dummies *}
+
+text {*
+  In most cases, defining a recursive function is just as simple as other definitions:
+*}
+
+fun fib :: "nat \<Rightarrow> nat"
+where
+  "fib 0 = 1"
+| "fib (Suc 0) = 1"
+| "fib (Suc (Suc n)) = fib n + fib (Suc n)"
+
+text {*
+  The syntax is rather self-explanatory: We introduce a function by
+  giving its name, its type, 
+  and a set of defining recursive equations.
+  If we leave out the type, the most general type will be
+  inferred, which can sometimes lead to surprises: Since both @{term
+  "1::nat"} and @{text "+"} are overloaded, we would end up
+  with @{text "fib :: nat \<Rightarrow> 'a::{one,plus}"}.
+*}
+
+text {*
+  The function always terminates, since its argument gets smaller in
+  every recursive call. 
+  Since HOL is a logic of total functions, termination is a
+  fundamental requirement to prevent inconsistencies\footnote{From the
+  \qt{definition} @{text "f(n) = f(n) + 1"} we could prove 
+  @{text "0 = 1"} by subtracting @{text "f(n)"} on both sides.}.
+  Isabelle tries to prove termination automatically when a definition
+  is made. In \S\ref{termination}, we will look at cases where this
+  fails and see what to do then.
+*}
+
+subsection {* Pattern matching *}
+
+text {* \label{patmatch}
+  Like in functional programming, we can use pattern matching to
+  define functions. At the moment we will only consider \emph{constructor
+  patterns}, which only consist of datatype constructors and
+  variables. Furthermore, patterns must be linear, i.e.\ all variables
+  on the left hand side of an equation must be distinct. In
+  \S\ref{genpats} we discuss more general pattern matching.
+
+  If patterns overlap, the order of the equations is taken into
+  account. The following function inserts a fixed element between any
+  two elements of a list:
+*}
+
+fun sep :: "'a \<Rightarrow> 'a list \<Rightarrow> 'a list"
+where
+  "sep a (x#y#xs) = x # a # sep a (y # xs)"
+| "sep a xs       = xs"
+
+text {* 
+  Overlapping patterns are interpreted as \qt{increments} to what is
+  already there: The second equation is only meant for the cases where
+  the first one does not match. Consequently, Isabelle replaces it
+  internally by the remaining cases, making the patterns disjoint:
+*}
+
+thm sep.simps
+
+text {* @{thm [display] sep.simps[no_vars]} *}
+
+text {* 
+  \noindent The equations from function definitions are automatically used in
+  simplification:
+*}
+
+lemma "sep 0 [1, 2, 3] = [1, 0, 2, 0, 3]"
+by simp
+
+subsection {* Induction *}
+
+text {*
+
+  Isabelle provides customized induction rules for recursive
+  functions. These rules follow the recursive structure of the
+  definition. Here is the rule @{text sep.induct} arising from the
+  above definition of @{const sep}:
+
+  @{thm [display] sep.induct}
+  
+  We have a step case for list with at least two elements, and two
+  base cases for the zero- and the one-element list. Here is a simple
+  proof about @{const sep} and @{const map}
+*}
+
+lemma "map f (sep x ys) = sep (f x) (map f ys)"
+apply (induct x ys rule: sep.induct)
+
+txt {*
+  We get three cases, like in the definition.
+
+  @{subgoals [display]}
+*}
+
+apply auto 
+done
+text {*
+
+  With the \cmd{fun} command, you can define about 80\% of the
+  functions that occur in practice. The rest of this tutorial explains
+  the remaining 20\%.
+*}
+
+
+section {* fun vs.\ function *}
+
+text {* 
+  The \cmd{fun} command provides a
+  convenient shorthand notation for simple function definitions. In
+  this mode, Isabelle tries to solve all the necessary proof obligations
+  automatically. If any proof fails, the definition is
+  rejected. This can either mean that the definition is indeed faulty,
+  or that the default proof procedures are just not smart enough (or
+  rather: not designed) to handle the definition.
+
+  By expanding the abbreviation to the more verbose \cmd{function} command, these proof obligations become visible and can be analyzed or
+  solved manually. The expansion from \cmd{fun} to \cmd{function} is as follows:
+
+\end{isamarkuptext}
+
+
+\[\left[\;\begin{minipage}{0.25\textwidth}\vspace{6pt}
+\cmd{fun} @{text "f :: \<tau>"}\\%
+\cmd{where}\\%
+\hspace*{2ex}{\it equations}\\%
+\hspace*{2ex}\vdots\vspace*{6pt}
+\end{minipage}\right]
+\quad\equiv\quad
+\left[\;\begin{minipage}{0.48\textwidth}\vspace{6pt}
+\cmd{function} @{text "("}\cmd{sequential}@{text ") f :: \<tau>"}\\%
+\cmd{where}\\%
+\hspace*{2ex}{\it equations}\\%
+\hspace*{2ex}\vdots\\%
+\cmd{by} @{text "pat_completeness auto"}\\%
+\cmd{termination by} @{text "lexicographic_order"}\vspace{6pt}
+\end{minipage}
+\right]\]
+
+\begin{isamarkuptext}
+  \vspace*{1em}
+  \noindent Some details have now become explicit:
+
+  \begin{enumerate}
+  \item The \cmd{sequential} option enables the preprocessing of
+  pattern overlaps which we already saw. Without this option, the equations
+  must already be disjoint and complete. The automatic completion only
+  works with constructor patterns.
+
+  \item A function definition produces a proof obligation which
+  expresses completeness and compatibility of patterns (we talk about
+  this later). The combination of the methods @{text "pat_completeness"} and
+  @{text "auto"} is used to solve this proof obligation.
+
+  \item A termination proof follows the definition, started by the
+  \cmd{termination} command. This will be explained in \S\ref{termination}.
+ \end{enumerate}
+  Whenever a \cmd{fun} command fails, it is usually a good idea to
+  expand the syntax to the more verbose \cmd{function} form, to see
+  what is actually going on.
+ *}
+
+
+section {* Termination *}
+
+text {*\label{termination}
+  The method @{text "lexicographic_order"} is the default method for
+  termination proofs. It can prove termination of a
+  certain class of functions by searching for a suitable lexicographic
+  combination of size measures. Of course, not all functions have such
+  a simple termination argument. For them, we can specify the termination
+  relation manually.
+*}
+
+subsection {* The {\tt relation} method *}
+text{*
+  Consider the following function, which sums up natural numbers up to
+  @{text "N"}, using a counter @{text "i"}:
+*}
+
+function sum :: "nat \<Rightarrow> nat \<Rightarrow> nat"
+where
+  "sum i N = (if i > N then 0 else i + sum (Suc i) N)"
+by pat_completeness auto
+
+text {*
+  \noindent The @{text "lexicographic_order"} method fails on this example, because none of the
+  arguments decreases in the recursive call, with respect to the standard size ordering.
+  To prove termination manually, we must provide a custom wellfounded relation.
+
+  The termination argument for @{text "sum"} is based on the fact that
+  the \emph{difference} between @{text "i"} and @{text "N"} gets
+  smaller in every step, and that the recursion stops when @{text "i"}
+  is greater than @{text "N"}. Phrased differently, the expression 
+  @{text "N + 1 - i"} always decreases.
+
+  We can use this expression as a measure function suitable to prove termination.
+*}
+
+termination sum
+apply (relation "measure (\<lambda>(i,N). N + 1 - i)")
+
+txt {*
+  The \cmd{termination} command sets up the termination goal for the
+  specified function @{text "sum"}. If the function name is omitted, it
+  implicitly refers to the last function definition.
+
+  The @{text relation} method takes a relation of
+  type @{typ "('a \<times> 'a) set"}, where @{typ "'a"} is the argument type of
+  the function. If the function has multiple curried arguments, then
+  these are packed together into a tuple, as it happened in the above
+  example.
+
+  The predefined function @{term[source] "measure :: ('a \<Rightarrow> nat) \<Rightarrow> ('a \<times> 'a) set"} constructs a
+  wellfounded relation from a mapping into the natural numbers (a
+  \emph{measure function}). 
+
+  After the invocation of @{text "relation"}, we must prove that (a)
+  the relation we supplied is wellfounded, and (b) that the arguments
+  of recursive calls indeed decrease with respect to the
+  relation:
+
+  @{subgoals[display,indent=0]}
+
+  These goals are all solved by @{text "auto"}:
+*}
+
+apply auto
+done
+
+text {*
+  Let us complicate the function a little, by adding some more
+  recursive calls: 
+*}
+
+function foo :: "nat \<Rightarrow> nat \<Rightarrow> nat"
+where
+  "foo i N = (if i > N 
+              then (if N = 0 then 0 else foo 0 (N - 1))
+              else i + foo (Suc i) N)"
+by pat_completeness auto
+
+text {*
+  When @{text "i"} has reached @{text "N"}, it starts at zero again
+  and @{text "N"} is decremented.
+  This corresponds to a nested
+  loop where one index counts up and the other down. Termination can
+  be proved using a lexicographic combination of two measures, namely
+  the value of @{text "N"} and the above difference. The @{const
+  "measures"} combinator generalizes @{text "measure"} by taking a
+  list of measure functions.  
+*}
+
+termination 
+by (relation "measures [\<lambda>(i, N). N, \<lambda>(i,N). N + 1 - i]") auto
+
+subsection {* How @{text "lexicographic_order"} works *}
+
+(*fun fails :: "nat \<Rightarrow> nat list \<Rightarrow> nat"
+where
+  "fails a [] = a"
+| "fails a (x#xs) = fails (x + a) (x # xs)"
+*)
+
+text {*
+  To see how the automatic termination proofs work, let's look at an
+  example where it fails\footnote{For a detailed discussion of the
+  termination prover, see \cite{bulwahnKN07}}:
+
+\end{isamarkuptext}  
+\cmd{fun} @{text "fails :: \"nat \<Rightarrow> nat list \<Rightarrow> nat\""}\\%
+\cmd{where}\\%
+\hspace*{2ex}@{text "\"fails a [] = a\""}\\%
+|\hspace*{1.5ex}@{text "\"fails a (x#xs) = fails (x + a) (x#xs)\""}\\
+\begin{isamarkuptext}
+
+\noindent Isabelle responds with the following error:
+
+\begin{isabelle}
+*** Unfinished subgoals:\newline
+*** (a, 1, <):\newline
+*** \ 1.~@{text "\<And>x. x = 0"}\newline
+*** (a, 1, <=):\newline
+*** \ 1.~False\newline
+*** (a, 2, <):\newline
+*** \ 1.~False\newline
+*** Calls:\newline
+*** a) @{text "(a, x # xs) -->> (x + a, x # xs)"}\newline
+*** Measures:\newline
+*** 1) @{text "\<lambda>x. size (fst x)"}\newline
+*** 2) @{text "\<lambda>x. size (snd x)"}\newline
+*** Result matrix:\newline
+*** \ \ \ \ 1\ \ 2  \newline
+*** a:  ?   <= \newline
+*** Could not find lexicographic termination order.\newline
+*** At command "fun".\newline
+\end{isabelle}
+*}
+
+
+text {*
+  The key to this error message is the matrix at the bottom. The rows
+  of that matrix correspond to the different recursive calls (In our
+  case, there is just one). The columns are the function's arguments 
+  (expressed through different measure functions, which map the
+  argument tuple to a natural number). 
+
+  The contents of the matrix summarize what is known about argument
+  descents: The second argument has a weak descent (@{text "<="}) at the
+  recursive call, and for the first argument nothing could be proved,
+  which is expressed by @{text "?"}. In general, there are the values
+  @{text "<"}, @{text "<="} and @{text "?"}.
+
+  For the failed proof attempts, the unfinished subgoals are also
+  printed. Looking at these will often point to a missing lemma.
+
+%  As a more real example, here is quicksort:
+*}
+(*
+function qs :: "nat list \<Rightarrow> nat list"
+where
+  "qs [] = []"
+| "qs (x#xs) = qs [y\<in>xs. y < x] @ x # qs [y\<in>xs. y \<ge> x]"
+by pat_completeness auto
+
+termination apply lexicographic_order
+
+text {* If we try @{text "lexicographic_order"} method, we get the
+  following error *}
+termination by (lexicographic_order simp:l2)
+
+lemma l: "x \<le> y \<Longrightarrow> x < Suc y" by arith
+
+function 
+
+*)
+
+section {* Mutual Recursion *}
+
+text {*
+  If two or more functions call one another mutually, they have to be defined
+  in one step. Here are @{text "even"} and @{text "odd"}:
+*}
+
+function even :: "nat \<Rightarrow> bool"
+    and odd  :: "nat \<Rightarrow> bool"
+where
+  "even 0 = True"
+| "odd 0 = False"
+| "even (Suc n) = odd n"
+| "odd (Suc n) = even n"
+by pat_completeness auto
+
+text {*
+  To eliminate the mutual dependencies, Isabelle internally
+  creates a single function operating on the sum
+  type @{typ "nat + nat"}. Then, @{const even} and @{const odd} are
+  defined as projections. Consequently, termination has to be proved
+  simultaneously for both functions, by specifying a measure on the
+  sum type: 
+*}
+
+termination 
+by (relation "measure (\<lambda>x. case x of Inl n \<Rightarrow> n | Inr n \<Rightarrow> n)") auto
+
+text {* 
+  We could also have used @{text lexicographic_order}, which
+  supports mutual recursive termination proofs to a certain extent.
+*}
+
+subsection {* Induction for mutual recursion *}
+
+text {*
+
+  When functions are mutually recursive, proving properties about them
+  generally requires simultaneous induction. The induction rule @{text "even_odd.induct"}
+  generated from the above definition reflects this.
+
+  Let us prove something about @{const even} and @{const odd}:
+*}
+
+lemma even_odd_mod2:
+  "even n = (n mod 2 = 0)"
+  "odd n = (n mod 2 = 1)"
+
+txt {* 
+  We apply simultaneous induction, specifying the induction variable
+  for both goals, separated by \cmd{and}:  *}
+
+apply (induct n and n rule: even_odd.induct)
+
+txt {* 
+  We get four subgoals, which correspond to the clauses in the
+  definition of @{const even} and @{const odd}:
+  @{subgoals[display,indent=0]}
+  Simplification solves the first two goals, leaving us with two
+  statements about the @{text "mod"} operation to prove:
+*}
+
+apply simp_all
+
+txt {* 
+  @{subgoals[display,indent=0]} 
+
+  \noindent These can be handled by Isabelle's arithmetic decision procedures.
+  
+*}
+
+apply arith
+apply arith
+done
+
+text {*
+  In proofs like this, the simultaneous induction is really essential:
+  Even if we are just interested in one of the results, the other
+  one is necessary to strengthen the induction hypothesis. If we leave
+  out the statement about @{const odd} and just write @{term True} instead,
+  the same proof fails:
+*}
+
+lemma failed_attempt:
+  "even n = (n mod 2 = 0)"
+  "True"
+apply (induct n rule: even_odd.induct)
+
+txt {*
+  \noindent Now the third subgoal is a dead end, since we have no
+  useful induction hypothesis available:
+
+  @{subgoals[display,indent=0]} 
+*}
+
+oops
+
+section {* General pattern matching *}
+text{*\label{genpats} *}
+
+subsection {* Avoiding automatic pattern splitting *}
+
+text {*
+
+  Up to now, we used pattern matching only on datatypes, and the
+  patterns were always disjoint and complete, and if they weren't,
+  they were made disjoint automatically like in the definition of
+  @{const "sep"} in \S\ref{patmatch}.
+
+  This automatic splitting can significantly increase the number of
+  equations involved, and this is not always desirable. The following
+  example shows the problem:
+  
+  Suppose we are modeling incomplete knowledge about the world by a
+  three-valued datatype, which has values @{term "T"}, @{term "F"}
+  and @{term "X"} for true, false and uncertain propositions, respectively. 
+*}
+
+datatype P3 = T | F | X
+
+text {* \noindent Then the conjunction of such values can be defined as follows: *}
+
+fun And :: "P3 \<Rightarrow> P3 \<Rightarrow> P3"
+where
+  "And T p = p"
+| "And p T = p"
+| "And p F = F"
+| "And F p = F"
+| "And X X = X"
+
+
+text {* 
+  This definition is useful, because the equations can directly be used
+  as simplification rules. But the patterns overlap: For example,
+  the expression @{term "And T T"} is matched by both the first and
+  the second equation. By default, Isabelle makes the patterns disjoint by
+  splitting them up, producing instances:
+*}
+
+thm And.simps
+
+text {*
+  @{thm[indent=4] And.simps}
+  
+  \vspace*{1em}
+  \noindent There are several problems with this:
+
+  \begin{enumerate}
+  \item If the datatype has many constructors, there can be an
+  explosion of equations. For @{const "And"}, we get seven instead of
+  five equations, which can be tolerated, but this is just a small
+  example.
+
+  \item Since splitting makes the equations \qt{less general}, they
+  do not always match in rewriting. While the term @{term "And x F"}
+  can be simplified to @{term "F"} with the original equations, a
+  (manual) case split on @{term "x"} is now necessary.
+
+  \item The splitting also concerns the induction rule @{text
+  "And.induct"}. Instead of five premises it now has seven, which
+  means that our induction proofs will have more cases.
+
+  \item In general, it increases clarity if we get the same definition
+  back which we put in.
+  \end{enumerate}
+
+  If we do not want the automatic splitting, we can switch it off by
+  leaving out the \cmd{sequential} option. However, we will have to
+  prove that our pattern matching is consistent\footnote{This prevents
+  us from defining something like @{term "f x = True"} and @{term "f x
+  = False"} simultaneously.}:
+*}
+
+function And2 :: "P3 \<Rightarrow> P3 \<Rightarrow> P3"
+where
+  "And2 T p = p"
+| "And2 p T = p"
+| "And2 p F = F"
+| "And2 F p = F"
+| "And2 X X = X"
+
+txt {*
+  \noindent Now let's look at the proof obligations generated by a
+  function definition. In this case, they are:
+
+  @{subgoals[display,indent=0]}\vspace{-1.2em}\hspace{3cm}\vdots\vspace{1.2em}
+
+  The first subgoal expresses the completeness of the patterns. It has
+  the form of an elimination rule and states that every @{term x} of
+  the function's input type must match at least one of the patterns\footnote{Completeness could
+  be equivalently stated as a disjunction of existential statements: 
+@{term "(\<exists>p. x = (T, p)) \<or> (\<exists>p. x = (p, T)) \<or> (\<exists>p. x = (p, F)) \<or>
+  (\<exists>p. x = (F, p)) \<or> (x = (X, X))"}, and you can use the method @{text atomize_elim} to get that form instead.}. If the patterns just involve
+  datatypes, we can solve it with the @{text "pat_completeness"}
+  method:
+*}
+
+apply pat_completeness
+
+txt {*
+  The remaining subgoals express \emph{pattern compatibility}. We do
+  allow that an input value matches multiple patterns, but in this
+  case, the result (i.e.~the right hand sides of the equations) must
+  also be equal. For each pair of two patterns, there is one such
+  subgoal. Usually this needs injectivity of the constructors, which
+  is used automatically by @{text "auto"}.
+*}
+
+by auto
+
+
+subsection {* Non-constructor patterns *}
+
+text {*
+  Most of Isabelle's basic types take the form of inductive datatypes,
+  and usually pattern matching works on the constructors of such types. 
+  However, this need not be always the case, and the \cmd{function}
+  command handles other kind of patterns, too.
+
+  One well-known instance of non-constructor patterns are
+  so-called \emph{$n+k$-patterns}, which are a little controversial in
+  the functional programming world. Here is the initial fibonacci
+  example with $n+k$-patterns:
+*}
+
+function fib2 :: "nat \<Rightarrow> nat"
+where
+  "fib2 0 = 1"
+| "fib2 1 = 1"
+| "fib2 (n + 2) = fib2 n + fib2 (Suc n)"
+
+(*<*)ML_val "goals_limit := 1"(*>*)
+txt {*
+  This kind of matching is again justified by the proof of pattern
+  completeness and compatibility. 
+  The proof obligation for pattern completeness states that every natural number is
+  either @{term "0::nat"}, @{term "1::nat"} or @{term "n +
+  (2::nat)"}:
+
+  @{subgoals[display,indent=0]}
+
+  This is an arithmetic triviality, but unfortunately the
+  @{text arith} method cannot handle this specific form of an
+  elimination rule. However, we can use the method @{text
+  "atomize_elim"} to do an ad-hoc conversion to a disjunction of
+  existentials, which can then be solved by the arithmetic decision procedure.
+  Pattern compatibility and termination are automatic as usual.
+*}
+(*<*)ML_val "goals_limit := 10"(*>*)
+apply atomize_elim
+apply arith
+apply auto
+done
+termination by lexicographic_order
+text {*
+  We can stretch the notion of pattern matching even more. The
+  following function is not a sensible functional program, but a
+  perfectly valid mathematical definition:
+*}
+
+function ev :: "nat \<Rightarrow> bool"
+where
+  "ev (2 * n) = True"
+| "ev (2 * n + 1) = False"
+apply atomize_elim
+by arith+
+termination by (relation "{}") simp
+
+text {*
+  This general notion of pattern matching gives you a certain freedom
+  in writing down specifications. However, as always, such freedom should
+  be used with care:
+
+  If we leave the area of constructor
+  patterns, we have effectively departed from the world of functional
+  programming. This means that it is no longer possible to use the
+  code generator, and expect it to generate ML code for our
+  definitions. Also, such a specification might not work very well together with
+  simplification. Your mileage may vary.
+*}
+
+
+subsection {* Conditional equations *}
+
+text {* 
+  The function package also supports conditional equations, which are
+  similar to guards in a language like Haskell. Here is Euclid's
+  algorithm written with conditional patterns\footnote{Note that the
+  patterns are also overlapping in the base case}:
+*}
+
+function gcd :: "nat \<Rightarrow> nat \<Rightarrow> nat"
+where
+  "gcd x 0 = x"
+| "gcd 0 y = y"
+| "x < y \<Longrightarrow> gcd (Suc x) (Suc y) = gcd (Suc x) (y - x)"
+| "\<not> x < y \<Longrightarrow> gcd (Suc x) (Suc y) = gcd (x - y) (Suc y)"
+by (atomize_elim, auto, arith)
+termination by lexicographic_order
+
+text {*
+  By now, you can probably guess what the proof obligations for the
+  pattern completeness and compatibility look like. 
+
+  Again, functions with conditional patterns are not supported by the
+  code generator.
+*}
+
+
+subsection {* Pattern matching on strings *}
+
+text {*
+  As strings (as lists of characters) are normal datatypes, pattern
+  matching on them is possible, but somewhat problematic. Consider the
+  following definition:
+
+\end{isamarkuptext}
+\noindent\cmd{fun} @{text "check :: \"string \<Rightarrow> bool\""}\\%
+\cmd{where}\\%
+\hspace*{2ex}@{text "\"check (''good'') = True\""}\\%
+@{text "| \"check s = False\""}
+\begin{isamarkuptext}
+
+  \noindent An invocation of the above \cmd{fun} command does not
+  terminate. What is the problem? Strings are lists of characters, and
+  characters are a datatype with a lot of constructors. Splitting the
+  catch-all pattern thus leads to an explosion of cases, which cannot
+  be handled by Isabelle.
+
+  There are two things we can do here. Either we write an explicit
+  @{text "if"} on the right hand side, or we can use conditional patterns:
+*}
+
+function check :: "string \<Rightarrow> bool"
+where
+  "check (''good'') = True"
+| "s \<noteq> ''good'' \<Longrightarrow> check s = False"
+by auto
+
+
+section {* Partiality *}
+
+text {* 
+  In HOL, all functions are total. A function @{term "f"} applied to
+  @{term "x"} always has the value @{term "f x"}, and there is no notion
+  of undefinedness. 
+  This is why we have to do termination
+  proofs when defining functions: The proof justifies that the
+  function can be defined by wellfounded recursion.
+
+  However, the \cmd{function} package does support partiality to a
+  certain extent. Let's look at the following function which looks
+  for a zero of a given function f. 
+*}
+
+function (*<*)(domintros, tailrec)(*>*)findzero :: "(nat \<Rightarrow> nat) \<Rightarrow> nat \<Rightarrow> nat"
+where
+  "findzero f n = (if f n = 0 then n else findzero f (Suc n))"
+by pat_completeness auto
+(*<*)declare findzero.simps[simp del](*>*)
+
+text {*
+  \noindent Clearly, any attempt of a termination proof must fail. And without
+  that, we do not get the usual rules @{text "findzero.simps"} and 
+  @{text "findzero.induct"}. So what was the definition good for at all?
+*}
+
+subsection {* Domain predicates *}
+
+text {*
+  The trick is that Isabelle has not only defined the function @{const findzero}, but also
+  a predicate @{term "findzero_dom"} that characterizes the values where the function
+  terminates: the \emph{domain} of the function. If we treat a
+  partial function just as a total function with an additional domain
+  predicate, we can derive simplification and
+  induction rules as we do for total functions. They are guarded
+  by domain conditions and are called @{text psimps} and @{text
+  pinduct}: 
+*}
+
+text {*
+  \noindent\begin{minipage}{0.79\textwidth}@{thm[display,margin=85] findzero.psimps}\end{minipage}
+  \hfill(@{text "findzero.psimps"})
+  \vspace{1em}
+
+  \noindent\begin{minipage}{0.79\textwidth}@{thm[display,margin=85] findzero.pinduct}\end{minipage}
+  \hfill(@{text "findzero.pinduct"})
+*}
+
+text {*
+  Remember that all we
+  are doing here is use some tricks to make a total function appear
+  as if it was partial. We can still write the term @{term "findzero
+  (\<lambda>x. 1) 0"} and like any other term of type @{typ nat} it is equal
+  to some natural number, although we might not be able to find out
+  which one. The function is \emph{underdefined}.
+
+  But it is defined enough to prove something interesting about it. We
+  can prove that if @{term "findzero f n"}
+  terminates, it indeed returns a zero of @{term f}:
+*}
+
+lemma findzero_zero: "findzero_dom (f, n) \<Longrightarrow> f (findzero f n) = 0"
+
+txt {* \noindent We apply induction as usual, but using the partial induction
+  rule: *}
+
+apply (induct f n rule: findzero.pinduct)
+
+txt {* \noindent This gives the following subgoals:
+
+  @{subgoals[display,indent=0]}
+
+  \noindent The hypothesis in our lemma was used to satisfy the first premise in
+  the induction rule. However, we also get @{term
+  "findzero_dom (f, n)"} as a local assumption in the induction step. This
+  allows to unfold @{term "findzero f n"} using the @{text psimps}
+  rule, and the rest is trivial. Since the @{text psimps} rules carry the
+  @{text "[simp]"} attribute by default, we just need a single step:
+ *}
+apply simp
+done
+
+text {*
+  Proofs about partial functions are often not harder than for total
+  functions. Fig.~\ref{findzero_isar} shows a slightly more
+  complicated proof written in Isar. It is verbose enough to show how
+  partiality comes into play: From the partial induction, we get an
+  additional domain condition hypothesis. Observe how this condition
+  is applied when calls to @{term findzero} are unfolded.
+*}
+
+text_raw {*
+\begin{figure}
+\hrule\vspace{6pt}
+\begin{minipage}{0.8\textwidth}
+\isabellestyle{it}
+\isastyle\isamarkuptrue
+*}
+lemma "\<lbrakk>findzero_dom (f, n); x \<in> {n ..< findzero f n}\<rbrakk> \<Longrightarrow> f x \<noteq> 0"
+proof (induct rule: findzero.pinduct)
+  fix f n assume dom: "findzero_dom (f, n)"
+               and IH: "\<lbrakk>f n \<noteq> 0; x \<in> {Suc n ..< findzero f (Suc n)}\<rbrakk> \<Longrightarrow> f x \<noteq> 0"
+               and x_range: "x \<in> {n ..< findzero f n}"
+  have "f n \<noteq> 0"
+  proof 
+    assume "f n = 0"
+    with dom have "findzero f n = n" by simp
+    with x_range show False by auto
+  qed
+  
+  from x_range have "x = n \<or> x \<in> {Suc n ..< findzero f n}" by auto
+  thus "f x \<noteq> 0"
+  proof
+    assume "x = n"
+    with `f n \<noteq> 0` show ?thesis by simp
+  next
+    assume "x \<in> {Suc n ..< findzero f n}"
+    with dom and `f n \<noteq> 0` have "x \<in> {Suc n ..< findzero f (Suc n)}" by simp
+    with IH and `f n \<noteq> 0`
+    show ?thesis by simp
+  qed
+qed
+text_raw {*
+\isamarkupfalse\isabellestyle{tt}
+\end{minipage}\vspace{6pt}\hrule
+\caption{A proof about a partial function}\label{findzero_isar}
+\end{figure}
+*}
+
+subsection {* Partial termination proofs *}
+
+text {*
+  Now that we have proved some interesting properties about our
+  function, we should turn to the domain predicate and see if it is
+  actually true for some values. Otherwise we would have just proved
+  lemmas with @{term False} as a premise.
+
+  Essentially, we need some introduction rules for @{text
+  findzero_dom}. The function package can prove such domain
+  introduction rules automatically. But since they are not used very
+  often (they are almost never needed if the function is total), this
+  functionality is disabled by default for efficiency reasons. So we have to go
+  back and ask for them explicitly by passing the @{text
+  "(domintros)"} option to the function package:
+
+\vspace{1ex}
+\noindent\cmd{function} @{text "(domintros) findzero :: \"(nat \<Rightarrow> nat) \<Rightarrow> nat \<Rightarrow> nat\""}\\%
+\cmd{where}\isanewline%
+\ \ \ldots\\
+
+  \noindent Now the package has proved an introduction rule for @{text findzero_dom}:
+*}
+
+thm findzero.domintros
+
+text {*
+  @{thm[display] findzero.domintros}
+
+  Domain introduction rules allow to show that a given value lies in the
+  domain of a function, if the arguments of all recursive calls
+  are in the domain as well. They allow to do a \qt{single step} in a
+  termination proof. Usually, you want to combine them with a suitable
+  induction principle.
+
+  Since our function increases its argument at recursive calls, we
+  need an induction principle which works \qt{backwards}. We will use
+  @{text inc_induct}, which allows to do induction from a fixed number
+  \qt{downwards}:
+
+  \begin{center}@{thm inc_induct}\hfill(@{text "inc_induct"})\end{center}
+
+  Figure \ref{findzero_term} gives a detailed Isar proof of the fact
+  that @{text findzero} terminates if there is a zero which is greater
+  or equal to @{term n}. First we derive two useful rules which will
+  solve the base case and the step case of the induction. The
+  induction is then straightforward, except for the unusual induction
+  principle.
+
+*}
+
+text_raw {*
+\begin{figure}
+\hrule\vspace{6pt}
+\begin{minipage}{0.8\textwidth}
+\isabellestyle{it}
+\isastyle\isamarkuptrue
+*}
+lemma findzero_termination:
+  assumes "x \<ge> n" and "f x = 0"
+  shows "findzero_dom (f, n)"
+proof - 
+  have base: "findzero_dom (f, x)"
+    by (rule findzero.domintros) (simp add:`f x = 0`)
+
+  have step: "\<And>i. findzero_dom (f, Suc i) 
+    \<Longrightarrow> findzero_dom (f, i)"
+    by (rule findzero.domintros) simp
+
+  from `x \<ge> n` show ?thesis
+  proof (induct rule:inc_induct)
+    show "findzero_dom (f, x)" by (rule base)
+  next
+    fix i assume "findzero_dom (f, Suc i)"
+    thus "findzero_dom (f, i)" by (rule step)
+  qed
+qed      
+text_raw {*
+\isamarkupfalse\isabellestyle{tt}
+\end{minipage}\vspace{6pt}\hrule
+\caption{Termination proof for @{text findzero}}\label{findzero_term}
+\end{figure}
+*}
+      
+text {*
+  Again, the proof given in Fig.~\ref{findzero_term} has a lot of
+  detail in order to explain the principles. Using more automation, we
+  can also have a short proof:
+*}
+
+lemma findzero_termination_short:
+  assumes zero: "x >= n" 
+  assumes [simp]: "f x = 0"
+  shows "findzero_dom (f, n)"
+using zero
+by (induct rule:inc_induct) (auto intro: findzero.domintros)
+    
+text {*
+  \noindent It is simple to combine the partial correctness result with the
+  termination lemma:
+*}
+
+lemma findzero_total_correctness:
+  "f x = 0 \<Longrightarrow> f (findzero f 0) = 0"
+by (blast intro: findzero_zero findzero_termination)
+
+subsection {* Definition of the domain predicate *}
+
+text {*
+  Sometimes it is useful to know what the definition of the domain
+  predicate looks like. Actually, @{text findzero_dom} is just an
+  abbreviation:
+
+  @{abbrev[display] findzero_dom}
+
+  The domain predicate is the \emph{accessible part} of a relation @{const
+  findzero_rel}, which was also created internally by the function
+  package. @{const findzero_rel} is just a normal
+  inductive predicate, so we can inspect its definition by
+  looking at the introduction rules @{text findzero_rel.intros}.
+  In our case there is just a single rule:
+
+  @{thm[display] findzero_rel.intros}
+
+  The predicate @{const findzero_rel}
+  describes the \emph{recursion relation} of the function
+  definition. The recursion relation is a binary relation on
+  the arguments of the function that relates each argument to its
+  recursive calls. In general, there is one introduction rule for each
+  recursive call.
+
+  The predicate @{term "accp findzero_rel"} is the accessible part of
+  that relation. An argument belongs to the accessible part, if it can
+  be reached in a finite number of steps (cf.~its definition in @{text
+  "Wellfounded.thy"}).
+
+  Since the domain predicate is just an abbreviation, you can use
+  lemmas for @{const accp} and @{const findzero_rel} directly. Some
+  lemmas which are occasionally useful are @{text accpI}, @{text
+  accp_downward}, and of course the introduction and elimination rules
+  for the recursion relation @{text "findzero.intros"} and @{text "findzero.cases"}.
+*}
+
+(*lemma findzero_nicer_domintros:
+  "f x = 0 \<Longrightarrow> findzero_dom (f, x)"
+  "findzero_dom (f, Suc x) \<Longrightarrow> findzero_dom (f, x)"
+by (rule accpI, erule findzero_rel.cases, auto)+
+*)
+  
+subsection {* A Useful Special Case: Tail recursion *}
+
+text {*
+  The domain predicate is our trick that allows us to model partiality
+  in a world of total functions. The downside of this is that we have
+  to carry it around all the time. The termination proof above allowed
+  us to replace the abstract @{term "findzero_dom (f, n)"} by the more
+  concrete @{term "(x \<ge> n \<and> f x = (0::nat))"}, but the condition is still
+  there and can only be discharged for special cases.
+  In particular, the domain predicate guards the unfolding of our
+  function, since it is there as a condition in the @{text psimp}
+  rules. 
+
+  Now there is an important special case: We can actually get rid
+  of the condition in the simplification rules, \emph{if the function
+  is tail-recursive}. The reason is that for all tail-recursive
+  equations there is a total function satisfying them, even if they
+  are non-terminating. 
+
+%  A function is tail recursive, if each call to the function is either
+%  equal
+%
+%  So the outer form of the 
+%
+%if it can be written in the following
+%  form:
+%  {term[display] "f x = (if COND x then BASE x else f (LOOP x))"}
+
+
+  The function package internally does the right construction and can
+  derive the unconditional simp rules, if we ask it to do so. Luckily,
+  our @{const "findzero"} function is tail-recursive, so we can just go
+  back and add another option to the \cmd{function} command:
+
+\vspace{1ex}
+\noindent\cmd{function} @{text "(domintros, tailrec) findzero :: \"(nat \<Rightarrow> nat) \<Rightarrow> nat \<Rightarrow> nat\""}\\%
+\cmd{where}\isanewline%
+\ \ \ldots\\%
+
+  
+  \noindent Now, we actually get unconditional simplification rules, even
+  though the function is partial:
+*}
+
+thm findzero.simps
+
+text {*
+  @{thm[display] findzero.simps}
+
+  \noindent Of course these would make the simplifier loop, so we better remove
+  them from the simpset:
+*}
+
+declare findzero.simps[simp del]
+
+text {* 
+  Getting rid of the domain conditions in the simplification rules is
+  not only useful because it simplifies proofs. It is also required in
+  order to use Isabelle's code generator to generate ML code
+  from a function definition.
+  Since the code generator only works with equations, it cannot be
+  used with @{text "psimp"} rules. Thus, in order to generate code for
+  partial functions, they must be defined as a tail recursion.
+  Luckily, many functions have a relatively natural tail recursive
+  definition.
+*}
+
+section {* Nested recursion *}
+
+text {*
+  Recursive calls which are nested in one another frequently cause
+  complications, since their termination proof can depend on a partial
+  correctness property of the function itself. 
+
+  As a small example, we define the \qt{nested zero} function:
+*}
+
+function nz :: "nat \<Rightarrow> nat"
+where
+  "nz 0 = 0"
+| "nz (Suc n) = nz (nz n)"
+by pat_completeness auto
+
+text {*
+  If we attempt to prove termination using the identity measure on
+  naturals, this fails:
+*}
+
+termination
+  apply (relation "measure (\<lambda>n. n)")
+  apply auto
+
+txt {*
+  We get stuck with the subgoal
+
+  @{subgoals[display]}
+
+  Of course this statement is true, since we know that @{const nz} is
+  the zero function. And in fact we have no problem proving this
+  property by induction.
+*}
+(*<*)oops(*>*)
+lemma nz_is_zero: "nz_dom n \<Longrightarrow> nz n = 0"
+  by (induct rule:nz.pinduct) auto
+
+text {*
+  We formulate this as a partial correctness lemma with the condition
+  @{term "nz_dom n"}. This allows us to prove it with the @{text
+  pinduct} rule before we have proved termination. With this lemma,
+  the termination proof works as expected:
+*}
+
+termination
+  by (relation "measure (\<lambda>n. n)") (auto simp: nz_is_zero)
+
+text {*
+  As a general strategy, one should prove the statements needed for
+  termination as a partial property first. Then they can be used to do
+  the termination proof. This also works for less trivial
+  examples. Figure \ref{f91} defines the 91-function, a well-known
+  challenge problem due to John McCarthy, and proves its termination.
+*}
+
+text_raw {*
+\begin{figure}
+\hrule\vspace{6pt}
+\begin{minipage}{0.8\textwidth}
+\isabellestyle{it}
+\isastyle\isamarkuptrue
+*}
+
+function f91 :: "nat \<Rightarrow> nat"
+where
+  "f91 n = (if 100 < n then n - 10 else f91 (f91 (n + 11)))"
+by pat_completeness auto
+
+lemma f91_estimate: 
+  assumes trm: "f91_dom n" 
+  shows "n < f91 n + 11"
+using trm by induct auto
+
+termination
+proof
+  let ?R = "measure (\<lambda>x. 101 - x)"
+  show "wf ?R" ..
+
+  fix n :: nat assume "\<not> 100 < n" -- "Assumptions for both calls"
+
+  thus "(n + 11, n) \<in> ?R" by simp -- "Inner call"
+
+  assume inner_trm: "f91_dom (n + 11)" -- "Outer call"
+  with f91_estimate have "n + 11 < f91 (n + 11) + 11" .
+  with `\<not> 100 < n` show "(f91 (n + 11), n) \<in> ?R" by simp
+qed
+
+text_raw {*
+\isamarkupfalse\isabellestyle{tt}
+\end{minipage}
+\vspace{6pt}\hrule
+\caption{McCarthy's 91-function}\label{f91}
+\end{figure}
+*}
+
+
+section {* Higher-Order Recursion *}
+
+text {*
+  Higher-order recursion occurs when recursive calls
+  are passed as arguments to higher-order combinators such as @{const
+  map}, @{term filter} etc.
+  As an example, imagine a datatype of n-ary trees:
+*}
+
+datatype 'a tree = 
+  Leaf 'a 
+| Branch "'a tree list"
+
+
+text {* \noindent We can define a function which swaps the left and right subtrees recursively, using the 
+  list functions @{const rev} and @{const map}: *}
+
+fun mirror :: "'a tree \<Rightarrow> 'a tree"
+where
+  "mirror (Leaf n) = Leaf n"
+| "mirror (Branch l) = Branch (rev (map mirror l))"
+
+text {*
+  Although the definition is accepted without problems, let us look at the termination proof:
+*}
+
+termination proof
+  txt {*
+
+  As usual, we have to give a wellfounded relation, such that the
+  arguments of the recursive calls get smaller. But what exactly are
+  the arguments of the recursive calls when mirror is given as an
+  argument to @{const map}? Isabelle gives us the
+  subgoals
+
+  @{subgoals[display,indent=0]} 
+
+  So the system seems to know that @{const map} only
+  applies the recursive call @{term "mirror"} to elements
+  of @{term "l"}, which is essential for the termination proof.
+
+  This knowledge about @{const map} is encoded in so-called congruence rules,
+  which are special theorems known to the \cmd{function} command. The
+  rule for @{const map} is
+
+  @{thm[display] map_cong}
+
+  You can read this in the following way: Two applications of @{const
+  map} are equal, if the list arguments are equal and the functions
+  coincide on the elements of the list. This means that for the value 
+  @{term "map f l"} we only have to know how @{term f} behaves on
+  the elements of @{term l}.
+
+  Usually, one such congruence rule is
+  needed for each higher-order construct that is used when defining
+  new functions. In fact, even basic functions like @{const
+  If} and @{const Let} are handled by this mechanism. The congruence
+  rule for @{const If} states that the @{text then} branch is only
+  relevant if the condition is true, and the @{text else} branch only if it
+  is false:
+
+  @{thm[display] if_cong}
+  
+  Congruence rules can be added to the
+  function package by giving them the @{term fundef_cong} attribute.
+
+  The constructs that are predefined in Isabelle, usually
+  come with the respective congruence rules.
+  But if you define your own higher-order functions, you may have to
+  state and prove the required congruence rules yourself, if you want to use your
+  functions in recursive definitions. 
+*}
+(*<*)oops(*>*)
+
+subsection {* Congruence Rules and Evaluation Order *}
+
+text {* 
+  Higher order logic differs from functional programming languages in
+  that it has no built-in notion of evaluation order. A program is
+  just a set of equations, and it is not specified how they must be
+  evaluated. 
+
+  However for the purpose of function definition, we must talk about
+  evaluation order implicitly, when we reason about termination.
+  Congruence rules express that a certain evaluation order is
+  consistent with the logical definition. 
+
+  Consider the following function.
+*}
+
+function f :: "nat \<Rightarrow> bool"
+where
+  "f n = (n = 0 \<or> f (n - 1))"
+(*<*)by pat_completeness auto(*>*)
+
+text {*
+  For this definition, the termination proof fails. The default configuration
+  specifies no congruence rule for disjunction. We have to add a
+  congruence rule that specifies left-to-right evaluation order:
+
+  \vspace{1ex}
+  \noindent @{thm disj_cong}\hfill(@{text "disj_cong"})
+  \vspace{1ex}
+
+  Now the definition works without problems. Note how the termination
+  proof depends on the extra condition that we get from the congruence
+  rule.
+
+  However, as evaluation is not a hard-wired concept, we
+  could just turn everything around by declaring a different
+  congruence rule. Then we can make the reverse definition:
+*}
+
+lemma disj_cong2[fundef_cong]: 
+  "(\<not> Q' \<Longrightarrow> P = P') \<Longrightarrow> (Q = Q') \<Longrightarrow> (P \<or> Q) = (P' \<or> Q')"
+  by blast
+
+fun f' :: "nat \<Rightarrow> bool"
+where
+  "f' n = (f' (n - 1) \<or> n = 0)"
+
+text {*
+  \noindent These examples show that, in general, there is no \qt{best} set of
+  congruence rules.
+
+  However, such tweaking should rarely be necessary in
+  practice, as most of the time, the default set of congruence rules
+  works well.
+*}
+
+end