(* Title: Doc/Functions/Functions.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 @{thm [source] 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.
*}
subsection {* The @{text size_change} method *}
text {*
Some termination goals that are beyond the powers of
@{text lexicographic_order} can be solved automatically by the
more powerful @{text size_change} method, which uses a variant of
the size-change principle, together with some other
techniques. While the details are discussed
elsewhere\cite{krauss_phd},
here are a few typical situations where
@{text lexicographic_order} has difficulties and @{text size_change}
may be worth a try:
\begin{itemize}
\item Arguments are permuted in a recursive call.
\item Several mutually recursive functions with multiple arguments.
\item Unusual control flow (e.g., when some recursive calls cannot
occur in sequence).
\end{itemize}
Loading the theory @{text Multiset} makes the @{text size_change}
method a bit stronger: it can then use multiset orders internally.
*}
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 @{thm [source] "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 {* Elimination *}
text {*
A definition of function @{text f} gives rise to two kinds of elimination rules. Rule @{text f.cases}
simply describes case analysis according to the patterns used in the definition:
*}
fun list_to_option :: "'a list \<Rightarrow> 'a option"
where
"list_to_option [x] = Some x"
| "list_to_option _ = None"
thm list_to_option.cases
text {*
@{thm[display] list_to_option.cases}
Note that this rule does not mention the function at all, but only describes the cases used for
defining it. In contrast, the rule @{thm[source] list_to_option.elims} also tell us what the function
value will be in each case:
*}
thm list_to_option.elims
text {*
@{thm[display] list_to_option.elims}
\noindent
This lets us eliminate an assumption of the form @{prop "list_to_option xs = y"} and replace it
with the two cases, e.g.:
*}
lemma "list_to_option xs = y \<Longrightarrow> P"
proof (erule list_to_option.elims)
fix x assume "xs = [x]" "y = Some x" thus P sorry
next
assume "xs = []" "y = None" thus P sorry
next
fix a b xs' assume "xs = a # b # xs'" "y = None" thus P sorry
qed
text {*
Sometimes it is convenient to derive specialized versions of the @{text elim} rules above and
keep them around as facts explicitly. For example, it is natural to show that if
@{prop "list_to_option xs = Some y"}, then @{term xs} must be a singleton. The command
\cmd{fun\_cases} derives such facts automatically, by instantiating and simplifying the general
elimination rules given some pattern:
*}
fun_cases list_to_option_SomeE[elim]: "list_to_option xs = Some y"
thm list_to_option_SomeE
text {*
@{thm[display] list_to_option_SomeE}
*}
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 @{thm [source]
"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
termination by (relation "{}") simp
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)"
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,goals_limit=1]}
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.
*}
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
termination by (relation "{}") simp
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)(*>*)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
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(@{thm [source] "findzero.psimps"})
\vspace{1em}
\noindent\begin{minipage}{0.79\textwidth}@{thm[display,margin=85] findzero.pinduct}\end{minipage}
\hfill(@{thm [source] "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 unfolding @{term "findzero f n"} using the @{text psimps}
rule, and the rest is trivial.
*}
apply (simp add: findzero.psimps)
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 add: findzero.psimps)
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 add: findzero.psimps)
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
@{thm [source] inc_induct}, which allows to do induction from a fixed number
\qt{downwards}:
\begin{center}@{thm inc_induct}\hfill(@{thm [source] "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 @{thm [source] 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 @{thm [source] accpI}, @{thm [source]
accp_downward}, and of course the introduction and elimination rules
for the recursion relation @{thm [source] "findzero_rel.intros"} and @{thm
[source] "findzero_rel.cases"}.
*}
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 simp: nz.psimps)
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 simp: f91.psimps)
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(@{thm [source] "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