doc-src/Nitpick/nitpick.tex

added support for nonstandard models to Nitpick (based on an idea by Koen Claessen) and did other fixes to Nitpick

\documentclass[a4paper,12pt]{article}
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%\def\lbrakk{\mathopen{\lbrack\mkern-3.25mu\lbrack}}
%\def\rbrakk{\mathclose{\rbrack\mkern-3.255mu\rbrack}}
\def\lparr{\mathopen{(\mkern-4mu\mid}}
\def\rparr{\mathclose{\mid\mkern-4mu)}}

\def\unk{{?}}
\def\undef{(\lambda x.\; \unk)}
%\def\unr{\textit{others}}
\def\unr{\ldots}
\def\Abs#1{\hbox{\rm{\flqq}}{\,#1\,}\hbox{\rm{\frqq}}}
\def\Q{{\smash{\lower.2ex\hbox{$\scriptstyle?$}}}}

\hyphenation{Mini-Sat size-change First-Steps grand-parent nit-pick
counter-example counter-examples data-type data-types co-data-type 
co-data-types in-duc-tive co-in-duc-tive}

\urlstyle{tt}

\begin{document}

\selectlanguage{english}

\title{\includegraphics[scale=0.5]{isabelle_nitpick} \\[4ex]
Picking Nits \\[\smallskipamount]
\Large A User's Guide to Nitpick for Isabelle/HOL}
\author{\hbox{} \\
Jasmin Christian Blanchette \\
{\normalsize Institut f\"ur Informatik, Technische Universit\"at M\"unchen} \\
\hbox{}}

\maketitle

\tableofcontents

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\def\prew{\pre\advance\rightskip by-\leftmargin}
\def\postw{\post}

\section{Introduction}
\label{introduction}

Nitpick \cite{blanchette-nipkow-2009} is a counterexample generator for
Isabelle/HOL \cite{isa-tutorial} that is designed to handle formulas
combining (co)in\-duc\-tive datatypes, (co)in\-duc\-tively defined predicates, and
quantifiers. It builds on Kodkod \cite{torlak-jackson-2007}, a highly optimized
first-order relational model finder developed by the Software Design Group at
MIT. It is conceptually similar to Refute \cite{weber-2008}, from which it
borrows many ideas and code fragments, but it benefits from Kodkod's
optimizations and a new encoding scheme. The name Nitpick is shamelessly
appropriated from a now retired Alloy precursor.

Nitpick is easy to use---you simply enter \textbf{nitpick} after a putative
theorem and wait a few seconds. Nonetheless, there are situations where knowing
how it works under the hood and how it reacts to various options helps
increase the test coverage. This manual also explains how to install the tool on
your workstation. Should the motivation fail you, think of the many hours of
hard work Nitpick will save you. Proving non-theorems is \textsl{hard work}.

Another common use of Nitpick is to find out whether the axioms of a locale are
satisfiable, while the locale is being developed. To check this, it suffices to
write

\prew
\textbf{lemma}~``$\textit{False}$'' \\
\textbf{nitpick}~[\textit{show\_all}]
\postw

after the locale's \textbf{begin} keyword. To falsify \textit{False}, Nitpick
must find a model for the axioms. If it finds no model, we have an indication
that the axioms might be unsatisfiable.

Nitpick requires the Kodkodi package for Isabelle as well as a Java 1.5 virtual
machine called \texttt{java}. The examples presented in this manual can be found
in Isabelle's \texttt{src/HOL/Nitpick\_Examples/Manual\_Nits.thy} theory.

Throughout this manual, we will explicitly invoke the \textbf{nitpick} command.
Nitpick also provides an automatic mode that can be enabled using the
``Auto Nitpick'' option from the ``Isabelle'' menu in Proof General. In this
mode, Nitpick is run on every newly entered theorem, much like Auto Quickcheck.
The collective time limit for Auto Nitpick and Auto Quickcheck can be set using
the ``Auto Counterexample Time Limit'' option.

\newbox\boxA
\setbox\boxA=\hbox{\texttt{nospam}}

The known bugs and limitations at the time of writing are listed in
\S\ref{known-bugs-and-limitations}. Comments and bug reports concerning Nitpick
or this manual should be directed to
\texttt{blan{\color{white}nospam}\kern-\wd\boxA{}chette@\allowbreak
in.\allowbreak tum.\allowbreak de}.

\vskip2.5\smallskipamount

\textbf{Acknowledgment.} The author would like to thank Mark Summerfield for
suggesting several textual improvements.
% and Perry James for reporting a typo.

\section{First Steps}
\label{first-steps}

This section introduces Nitpick by presenting small examples. If possible, you
should try out the examples on your workstation. Your theory file should start
the standard way:

\prew
\textbf{theory}~\textit{Scratch} \\
\textbf{imports}~\textit{Main} \\
\textbf{begin}
\postw

The results presented here were obtained using the JNI version of MiniSat and
with multithreading disabled to reduce nondeterminism. This was done by adding
the line

\prew
\textbf{nitpick\_params} [\textit{sat\_solver}~= \textit{MiniSatJNI}, \,\textit{max\_threads}~= 1]
\postw

after the \textbf{begin} keyword. The JNI version of MiniSat is bundled with
Kodkodi and is precompiled for the major platforms. Other SAT solvers can also
be installed, as explained in \S\ref{optimizations}. If you have already
configured SAT solvers in Isabelle (e.g., for Refute), these will also be
available to Nitpick.

\subsection{Propositional Logic}
\label{propositional-logic}

Let's start with a trivial example from propositional logic:

\prew
\textbf{lemma}~``$P \longleftrightarrow Q$'' \\
\textbf{nitpick}
\postw

You should get the following output:

\prew
\slshape
Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $P = \textit{True}$ \\
\hbox{}\qquad\qquad $Q = \textit{False}$
\postw

Nitpick can also be invoked on individual subgoals, as in the example below:

\prew
\textbf{apply}~\textit{auto} \\[2\smallskipamount]
{\slshape goal (2 subgoals): \\
\phantom{0}1. $P\,\Longrightarrow\, Q$ \\
\phantom{0}2. $Q\,\Longrightarrow\, P$} \\[2\smallskipamount]
\textbf{nitpick}~1 \\[2\smallskipamount]
{\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $P = \textit{True}$ \\
\hbox{}\qquad\qquad $Q = \textit{False}$} \\[2\smallskipamount]
\textbf{nitpick}~2 \\[2\smallskipamount]
{\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $P = \textit{False}$ \\
\hbox{}\qquad\qquad $Q = \textit{True}$} \\[2\smallskipamount]
\textbf{oops}
\postw

\subsection{Type Variables}
\label{type-variables}

If you are left unimpressed by the previous example, don't worry. The next
one is more mind- and computer-boggling:

\prew
\textbf{lemma} ``$P~x\,\Longrightarrow\, P~(\textrm{THE}~y.\;P~y)$''
\postw
\pagebreak[2] %% TYPESETTING

The putative lemma involves the definite description operator, {THE}, presented
in section 5.10.1 of the Isabelle tutorial \cite{isa-tutorial}. The
operator is defined by the axiom $(\textrm{THE}~x.\; x = a) = a$. The putative
lemma is merely asserting the indefinite description operator axiom with {THE}
substituted for {SOME}.

The free variable $x$ and the bound variable $y$ have type $'a$. For formulas
containing type variables, Nitpick enumerates the possible domains for each type
variable, up to a given cardinality (8 by default), looking for a finite
countermodel:

\prew
\textbf{nitpick} [\textit{verbose}] \\[2\smallskipamount]
\slshape
Trying 8 scopes: \nopagebreak \\
\hbox{}\qquad \textit{card}~$'a$~= 1; \\
\hbox{}\qquad \textit{card}~$'a$~= 2; \\
\hbox{}\qquad $\qquad\vdots$ \\[.5\smallskipamount]
\hbox{}\qquad \textit{card}~$'a$~= 8. \\[2\smallskipamount]
Nitpick found a counterexample for \textit{card} $'a$~= 3: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $P = \{a_2,\, a_3\}$ \\
\hbox{}\qquad\qquad $x = a_3$ \\[2\smallskipamount]
Total time: 580 ms.
\postw

Nitpick found a counterexample in which $'a$ has cardinality 3. (For
cardinalities 1 and 2, the formula holds.) In the counterexample, the three
values of type $'a$ are written $a_1$, $a_2$, and $a_3$.

The message ``Trying $n$ scopes: {\ldots}''\ is shown only if the option
\textit{verbose} is enabled. You can specify \textit{verbose} each time you
invoke \textbf{nitpick}, or you can set it globally using the command

\prew
\textbf{nitpick\_params} [\textit{verbose}]
\postw

This command also displays the current default values for all of the options
supported by Nitpick. The options are listed in \S\ref{option-reference}.

\subsection{Constants}
\label{constants}

By just looking at Nitpick's output, it might not be clear why the
counterexample in \S\ref{type-variables} is genuine. Let's invoke Nitpick again,
this time telling it to show the values of the constants that occur in the
formula:

\prew
\textbf{lemma}~``$P~x\,\Longrightarrow\, P~(\textrm{THE}~y.\;P~y)$'' \\
\textbf{nitpick}~[\textit{show\_consts}] \\[2\smallskipamount]
\slshape
Nitpick found a counterexample for \textit{card} $'a$~= 3: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $P = \{a_2,\, a_3\}$ \\
\hbox{}\qquad\qquad $x = a_3$ \\
\hbox{}\qquad Constant: \nopagebreak \\
\hbox{}\qquad\qquad $\textit{The}~\textsl{fallback} = a_1$
\postw

We can see more clearly now. Since the predicate $P$ isn't true for a unique
value, $\textrm{THE}~y.\;P~y$ can denote any value of type $'a$, even
$a_1$. Since $P~a_1$ is false, the entire formula is falsified.

As an optimization, Nitpick's preprocessor introduced the special constant
``\textit{The} fallback'' corresponding to $\textrm{THE}~y.\;P~y$ (i.e.,
$\mathit{The}~(\lambda y.\;P~y)$) when there doesn't exist a unique $y$
satisfying $P~y$. We disable this optimization by passing the
\textit{full\_descrs} option:

\prew
\textbf{nitpick}~[\textit{full\_descrs},\, \textit{show\_consts}] \\[2\smallskipamount]
\slshape
Nitpick found a counterexample for \textit{card} $'a$~= 3: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $P = \{a_2,\, a_3\}$ \\
\hbox{}\qquad\qquad $x = a_3$ \\
\hbox{}\qquad Constant: \nopagebreak \\
\hbox{}\qquad\qquad $\hbox{\slshape THE}~y.\;P~y = a_1$
\postw

As the result of another optimization, Nitpick directly assigned a value to the
subterm $\textrm{THE}~y.\;P~y$, rather than to the \textit{The} constant. If we
disable this second optimization by using the command

\prew
\textbf{nitpick}~[\textit{dont\_specialize},\, \textit{full\_descrs},\,
\textit{show\_consts}]
\postw

we finally get \textit{The}:

\prew
\slshape Constant: \nopagebreak \\
\hbox{}\qquad $\mathit{The} = \undef{}
    (\!\begin{aligned}[t]%
    & \{\} := a_3,\> \{a_3\} := a_3,\> \{a_2\} := a_2, \\[-2pt] %% TYPESETTING
    & \{a_2, a_3\} := a_1,\> \{a_1\} := a_1,\> \{a_1, a_3\} := a_3, \\[-2pt]
    & \{a_1, a_2\} := a_3,\> \{a_1, a_2, a_3\} := a_3)\end{aligned}$
\postw

Notice that $\textit{The}~(\lambda y.\;P~y) = \textit{The}~\{a_2, a_3\} = a_1$,
just like before.\footnote{The Isabelle/HOL notation $f(x :=
y)$ denotes the function that maps $x$ to $y$ and that otherwise behaves like
$f$.}

Our misadventures with THE suggest adding `$\exists!x{.}$' (``there exists a
unique $x$ such that'') at the front of our putative lemma's assumption:

\prew
\textbf{lemma}~``$\exists {!}x.\; P~x\,\Longrightarrow\, P~(\textrm{THE}~y.\;P~y)$''
\postw

The fix appears to work:

\prew
\textbf{nitpick} \\[2\smallskipamount]
\slshape Nitpick found no counterexample.
\postw

We can further increase our confidence in the formula by exhausting all
cardinalities up to 50:

\prew
\textbf{nitpick} [\textit{card} $'a$~= 1--50]\footnote{The symbol `--'
can be entered as \texttt{-} (hyphen) or
\texttt{\char`\\\char`\<midarrow\char`\>}.} \\[2\smallskipamount]
\slshape Nitpick found no counterexample.
\postw

Let's see if Sledgehammer \cite{sledgehammer-2009} can find a proof:

\prew
\textbf{sledgehammer} \\[2\smallskipamount]
{\slshape Sledgehammer: external prover ``$e$'' for subgoal 1: \\
$\exists{!}x.\; P~x\,\Longrightarrow\, P~(\hbox{\slshape THE}~y.\; P~y)$ \\
Try this command: \textrm{apply}~(\textit{metis~the\_equality})} \\[2\smallskipamount]
\textbf{apply}~(\textit{metis~the\_equality\/}) \nopagebreak \\[2\smallskipamount]
{\slshape No subgoals!}% \\[2\smallskipamount]
%\textbf{done}
\postw

This must be our lucky day.

\subsection{Skolemization}
\label{skolemization}

Are all invertible functions onto? Let's find out:

\prew
\textbf{lemma} ``$\exists g.\; \forall x.~g~(f~x) = x
 \,\Longrightarrow\, \forall y.\; \exists x.~y = f~x$'' \\
\textbf{nitpick} \\[2\smallskipamount]
\slshape
Nitpick found a counterexample for \textit{card} $'a$~= 2 and \textit{card} $'b$~=~1: \\[2\smallskipamount]
\hbox{}\qquad Free variable: \nopagebreak \\
\hbox{}\qquad\qquad $f = \undef{}(b_1 := a_1)$ \\
\hbox{}\qquad Skolem constants: \nopagebreak \\
\hbox{}\qquad\qquad $g = \undef{}(a_1 := b_1,\> a_2 := b_1)$ \\
\hbox{}\qquad\qquad $y = a_2$
\postw

Although $f$ is the only free variable occurring in the formula, Nitpick also
displays values for the bound variables $g$ and $y$. These values are available
to Nitpick because it performs skolemization as a preprocessing step.

In the previous example, skolemization only affected the outermost quantifiers.
This is not always the case, as illustrated below:

\prew
\textbf{lemma} ``$\exists x.\; \forall f.\; f~x = x$'' \\
\textbf{nitpick} \\[2\smallskipamount]
\slshape
Nitpick found a counterexample for \textit{card} $'a$~= 2: \\[2\smallskipamount]
\hbox{}\qquad Skolem constant: \nopagebreak \\
\hbox{}\qquad\qquad $\lambda x.\; f =
    \undef{}(\!\begin{aligned}[t]
    & a_1 := \undef{}(a_1 := a_2,\> a_2 := a_1), \\[-2pt]
    & a_2 := \undef{}(a_1 := a_1,\> a_2 := a_1))\end{aligned}$
\postw

The variable $f$ is bound within the scope of $x$; therefore, $f$ depends on
$x$, as suggested by the notation $\lambda x.\,f$. If $x = a_1$, then $f$ is the
function that maps $a_1$ to $a_2$ and vice versa; otherwise, $x = a_2$ and $f$
maps both $a_1$ and $a_2$ to $a_1$. In both cases, $f~x \not= x$.

The source of the Skolem constants is sometimes more obscure:

\prew
\textbf{lemma} ``$\mathit{refl}~r\,\Longrightarrow\, \mathit{sym}~r$'' \\
\textbf{nitpick} \\[2\smallskipamount]
\slshape
Nitpick found a counterexample for \textit{card} $'a$~= 2: \\[2\smallskipamount]
\hbox{}\qquad Free variable: \nopagebreak \\
\hbox{}\qquad\qquad $r = \{(a_1, a_1),\, (a_2, a_1),\, (a_2, a_2)\}$ \\
\hbox{}\qquad Skolem constants: \nopagebreak \\
\hbox{}\qquad\qquad $\mathit{sym}.x = a_2$ \\
\hbox{}\qquad\qquad $\mathit{sym}.y = a_1$
\postw

What happened here is that Nitpick expanded the \textit{sym} constant to its
definition:

\prew
$\mathit{sym}~r \,\equiv\,
 \forall x\> y.\,\> (x, y) \in r \longrightarrow (y, x) \in r.$
\postw

As their names suggest, the Skolem constants $\mathit{sym}.x$ and
$\mathit{sym}.y$ are simply the bound variables $x$ and $y$
from \textit{sym}'s definition.

Although skolemization is a useful optimization, you can disable it by invoking
Nitpick with \textit{dont\_skolemize}. See \S\ref{optimizations} for details.

\subsection{Natural Numbers and Integers}
\label{natural-numbers-and-integers}

Because of the axiom of infinity, the type \textit{nat} does not admit any
finite models. To deal with this, Nitpick's approach is to consider finite
subsets $N$ of \textit{nat} and maps all numbers $\notin N$ to the undefined
value (displayed as `$\unk$'). The type \textit{int} is handled similarly.
Internally, undefined values lead to a three-valued logic.

Here is an example involving \textit{int}:

\prew
\textbf{lemma} ``$\lbrakk i \le j;\> n \le (m{\Colon}\mathit{int})\rbrakk \,\Longrightarrow\, i * n + j * m \le i * m + j * n$'' \\
\textbf{nitpick} \\[2\smallskipamount]
\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $i = 0$ \\
\hbox{}\qquad\qquad $j = 1$ \\
\hbox{}\qquad\qquad $m = 1$ \\
\hbox{}\qquad\qquad $n = 0$
\postw

Internally, Nitpick uses either a unary or a binary representation of numbers.
The unary representation is more efficient but only suitable for numbers very
close to zero. By default, Nitpick attempts to choose the more appropriate
encoding by inspecting the formula at hand. This behavior can be overridden by
passing either \textit{unary\_ints} or \textit{binary\_ints} as option. For
binary notation, the number of bits to use can be specified using
the \textit{bits} option. For example:

\prew
\textbf{nitpick} [\textit{binary\_ints}, \textit{bits}${} = 16$]
\postw

With infinite types, we don't always have the luxury of a genuine counterexample
and must often content ourselves with a potential one. The tedious task of
finding out whether the potential counterexample is in fact genuine can be
outsourced to \textit{auto} by passing \textit{check\_potential}. For example:

\prew
\textbf{lemma} ``$\forall n.\; \textit{Suc}~n \mathbin{\not=} n \,\Longrightarrow\, P$'' \\
\textbf{nitpick} [\textit{card~nat}~= 100, \textit{check\_potential}] \\[2\smallskipamount]
\slshape Nitpick found a potential counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variable: \nopagebreak \\
\hbox{}\qquad\qquad $P = \textit{False}$ \\[2\smallskipamount]
Confirmation by ``\textit{auto}'': The above counterexample is genuine.
\postw

You might wonder why the counterexample is first reported as potential. The root
of the problem is that the bound variable in $\forall n.\; \textit{Suc}~n
\mathbin{\not=} n$ ranges over an infinite type. If Nitpick finds an $n$ such
that $\textit{Suc}~n \mathbin{=} n$, it evaluates the assumption to
\textit{False}; but otherwise, it does not know anything about values of $n \ge
\textit{card~nat}$ and must therefore evaluate the assumption to $\unk$, not
\textit{True}. Since the assumption can never be satisfied, the putative lemma
can never be falsified.

Incidentally, if you distrust the so-called genuine counterexamples, you can
enable \textit{check\_\allowbreak genuine} to verify them as well. However, be
aware that \textit{auto} will usually fail to prove that the counterexample is
genuine or spurious.

Some conjectures involving elementary number theory make Nitpick look like a
giant with feet of clay:

\prew
\textbf{lemma} ``$P~\textit{Suc}$'' \\
\textbf{nitpick} [\textit{card} = 1--6] \\[2\smallskipamount]
\slshape
Nitpick found no counterexample.
\postw

On any finite set $N$, \textit{Suc} is a partial function; for example, if $N =
\{0, 1, \ldots, k\}$, then \textit{Suc} is $\{0 \mapsto 1,\, 1 \mapsto 2,\,
\ldots,\, k \mapsto \unk\}$, which evaluates to $\unk$ when passed as
argument to $P$. As a result, $P~\textit{Suc}$ is always $\unk$. The next
example is similar:

\prew
\textbf{lemma} ``$P~(\textit{op}~{+}\Colon
\textit{nat}\mathbin{\Rightarrow}\textit{nat}\mathbin{\Rightarrow}\textit{nat})$'' \\
\textbf{nitpick} [\textit{card nat} = 1] \\[2\smallskipamount]
{\slshape Nitpick found a counterexample:} \\[2\smallskipamount]
\hbox{}\qquad Free variable: \nopagebreak \\
\hbox{}\qquad\qquad $P = \{\}$ \\[2\smallskipamount]
\textbf{nitpick} [\textit{card nat} = 2] \\[2\smallskipamount]
{\slshape Nitpick found no counterexample.}
\postw

The problem here is that \textit{op}~+ is total when \textit{nat} is taken to be
$\{0\}$ but becomes partial as soon as we add $1$, because $1 + 1 \notin \{0,
1\}$.

Because numbers are infinite and are approximated using a three-valued logic,
there is usually no need to systematically enumerate domain sizes. If Nitpick
cannot find a genuine counterexample for \textit{card~nat}~= $k$, it is very
unlikely that one could be found for smaller domains. (The $P~(\textit{op}~{+})$
example above is an exception to this principle.) Nitpick nonetheless enumerates
all cardinalities from 1 to 8 for \textit{nat}, mainly because smaller
cardinalities are fast to handle and give rise to simpler counterexamples. This
is explained in more detail in \S\ref{scope-monotonicity}.

\subsection{Inductive Datatypes}
\label{inductive-datatypes}

Like natural numbers and integers, inductive datatypes with recursive
constructors admit no finite models and must be approximated by a subterm-closed
subset. For example, using a cardinality of 10 for ${'}a~\textit{list}$,
Nitpick looks for all counterexamples that can be built using at most 10
different lists.

Let's see with an example involving \textit{hd} (which returns the first element
of a list) and $@$ (which concatenates two lists):

\prew
\textbf{lemma} ``$\textit{hd}~(\textit{xs} \mathbin{@} [y, y]) = \textit{hd}~\textit{xs}$'' \\
\textbf{nitpick} \\[2\smallskipamount]
\slshape Nitpick found a counterexample for \textit{card} $'a$~= 3: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $\textit{xs} = []$ \\
\hbox{}\qquad\qquad $\textit{y} = a_3$
\postw

To see why the counterexample is genuine, we enable \textit{show\_consts}
and \textit{show\_\allowbreak datatypes}:

\prew
{\slshape Datatype:} \\
\hbox{}\qquad $'a$~\textit{list}~= $\{[],\, [a_3, a_3],\, [a_3],\, \unr\}$ \\
{\slshape Constants:} \\
\hbox{}\qquad $\lambda x_1.\; x_1 \mathbin{@} [y, y] = \undef([] := [a_3, a_3])$ \\
\hbox{}\qquad $\textit{hd} = \undef([] := a_2,\> [a_3, a_3] := a_3,\> [a_3] := a_3)$
\postw

Since $\mathit{hd}~[]$ is undefined in the logic, it may be given any value,
including $a_2$.

The second constant, $\lambda x_1.\; x_1 \mathbin{@} [y, y]$, is simply the
append operator whose second argument is fixed to be $[y, y]$. Appending $[a_3,
a_3]$ to $[a_3]$ would normally give $[a_3, a_3, a_3]$, but this value is not
representable in the subset of $'a$~\textit{list} considered by Nitpick, which
is shown under the ``Datatype'' heading; hence the result is $\unk$. Similarly,
appending $[a_3, a_3]$ to itself gives $\unk$.

Given \textit{card}~$'a = 3$ and \textit{card}~$'a~\textit{list} = 3$, Nitpick
considers the following subsets:

\kern-.5\smallskipamount %% TYPESETTING

\prew
\begin{multicols}{3}
$\{[],\, [a_1],\, [a_2]\}$; \\
$\{[],\, [a_1],\, [a_3]\}$; \\
$\{[],\, [a_2],\, [a_3]\}$; \\
$\{[],\, [a_1],\, [a_1, a_1]\}$; \\
$\{[],\, [a_1],\, [a_2, a_1]\}$; \\
$\{[],\, [a_1],\, [a_3, a_1]\}$; \\
$\{[],\, [a_2],\, [a_1, a_2]\}$; \\
$\{[],\, [a_2],\, [a_2, a_2]\}$; \\
$\{[],\, [a_2],\, [a_3, a_2]\}$; \\
$\{[],\, [a_3],\, [a_1, a_3]\}$; \\
$\{[],\, [a_3],\, [a_2, a_3]\}$; \\
$\{[],\, [a_3],\, [a_3, a_3]\}$.
\end{multicols}
\postw

\kern-2\smallskipamount %% TYPESETTING

All subterm-closed subsets of $'a~\textit{list}$ consisting of three values
are listed and only those. As an example of a non-subterm-closed subset,
consider $\mathcal{S} = \{[],\, [a_1],\,\allowbreak [a_1, a_3]\}$, and observe
that $[a_1, a_3]$ (i.e., $a_1 \mathbin{\#} [a_3]$) has $[a_3] \notin
\mathcal{S}$ as a subterm.

Here's another m\"ochtegern-lemma that Nitpick can refute without a blink:

\prew
\textbf{lemma} ``$\lbrakk \textit{length}~\textit{xs} = 1;\> \textit{length}~\textit{ys} = 1
\rbrakk \,\Longrightarrow\, \textit{xs} = \textit{ys}$''
\\
\textbf{nitpick} [\textit{show\_datatypes}] \\[2\smallskipamount]
\slshape Nitpick found a counterexample for \textit{card} $'a$~= 3: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $\textit{xs} = [a_2]$ \\
\hbox{}\qquad\qquad $\textit{ys} = [a_3]$ \\
\hbox{}\qquad Datatypes: \\
\hbox{}\qquad\qquad $\textit{nat} = \{0,\, 1,\, 2,\, \unr\}$ \\
\hbox{}\qquad\qquad $'a$~\textit{list} = $\{[],\, [a_3],\, [a_2],\, \unr\}$
\postw

Because datatypes are approximated using a three-valued logic, there is usually
no need to systematically enumerate cardinalities: If Nitpick cannot find a
genuine counterexample for \textit{card}~$'a~\textit{list}$~= 10, it is very
unlikely that one could be found for smaller cardinalities.

\subsection{Typedefs, Records, Rationals, and Reals}
\label{typedefs-records-rationals-and-reals}

Nitpick generally treats types declared using \textbf{typedef} as datatypes
whose single constructor is the corresponding \textit{Abs\_\kern.1ex} function.
For example:

\prew
\textbf{typedef}~\textit{three} = ``$\{0\Colon\textit{nat},\, 1,\, 2\}$'' \\
\textbf{by}~\textit{blast} \\[2\smallskipamount]
\textbf{definition}~$A \mathbin{\Colon} \textit{three}$ \textbf{where} ``\kern-.1em$A \,\equiv\, \textit{Abs\_\allowbreak three}~0$'' \\
\textbf{definition}~$B \mathbin{\Colon} \textit{three}$ \textbf{where} ``$B \,\equiv\, \textit{Abs\_three}~1$'' \\
\textbf{definition}~$C \mathbin{\Colon} \textit{three}$ \textbf{where} ``$C \,\equiv\, \textit{Abs\_three}~2$'' \\[2\smallskipamount]
\textbf{lemma} ``$\lbrakk P~A;\> P~B\rbrakk \,\Longrightarrow\, P~x$'' \\
\textbf{nitpick} [\textit{show\_datatypes}] \\[2\smallskipamount]
\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $P = \{\Abs{1},\, \Abs{0}\}$ \\
\hbox{}\qquad\qquad $x = \Abs{2}$ \\
\hbox{}\qquad Datatypes: \\
\hbox{}\qquad\qquad $\textit{nat} = \{0,\, 1,\, 2,\, \unr\}$ \\
\hbox{}\qquad\qquad $\textit{three} = \{\Abs{2},\, \Abs{1},\, \Abs{0},\, \unr\}$
\postw

%% MARK
In the output above, $\Abs{n}$ abbreviates $\textit{Abs\_three}~n$.

%% MARK
Records, which are implemented as \textbf{typedef}s behind the scenes, are
handled in much the same way:

\prew
\textbf{record} \textit{point} = \\
\hbox{}\quad $\textit{Xcoord} \mathbin{\Colon} \textit{int}$ \\
\hbox{}\quad $\textit{Ycoord} \mathbin{\Colon} \textit{int}$ \\[2\smallskipamount]
\textbf{lemma} ``$\textit{Xcoord}~(p\Colon\textit{point}) = \textit{Xcoord}~(q\Colon\textit{point})$'' \\
\textbf{nitpick} [\textit{show\_datatypes}] \\[2\smallskipamount]
\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $p = \lparr\textit{Xcoord} = 0,\> \textit{Ycoord} = 0\rparr$ \\
\hbox{}\qquad\qquad $q = \lparr\textit{Xcoord} = 1,\> \textit{Ycoord} = 1\rparr$ \\
\hbox{}\qquad Datatypes: \\
\hbox{}\qquad\qquad $\textit{int} = \{0,\, 1,\, \unr\}$ \\
\hbox{}\qquad\qquad $\textit{point} = \{\lparr\textit{Xcoord} = 1,\>
\textit{Ycoord} = 1\rparr,\> \lparr\textit{Xcoord} = 0,\> \textit{Ycoord} = 0\rparr,\, \unr\}$\kern-1pt %% QUIET
\postw

Finally, Nitpick provides rudimentary support for rationals and reals using a
similar approach:

\prew
\textbf{lemma} ``$4 * x + 3 * (y\Colon\textit{real}) \not= 1/2$'' \\
\textbf{nitpick} [\textit{show\_datatypes}] \\[2\smallskipamount]
\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $x = 1/2$ \\
\hbox{}\qquad\qquad $y = -1/2$ \\
\hbox{}\qquad Datatypes: \\
\hbox{}\qquad\qquad $\textit{nat} = \{0,\, 1,\, 2,\, 3,\, 4,\, 5,\, 6,\, 7,\, \unr\}$ \\
\hbox{}\qquad\qquad $\textit{int} = \{0,\, 1,\, 2,\, 3,\, 4,\, -3,\, -2,\, -1,\, \unr\}$ \\
\hbox{}\qquad\qquad $\textit{real} = \{1,\, 0,\, 4,\, -3/2,\, 3,\, 2,\, 1/2,\, -1/2,\, \unr\}$
\postw

\subsection{Inductive and Coinductive Predicates}
\label{inductive-and-coinductive-predicates}

Inductively defined predicates (and sets) are particularly problematic for
counterexample generators. They can make Quickcheck~\cite{berghofer-nipkow-2004}
loop forever and Refute~\cite{weber-2008} run out of resources. The crux of
the problem is that they are defined using a least fixed point construction.

Nitpick's philosophy is that not all inductive predicates are equal. Consider
the \textit{even} predicate below:

\prew
\textbf{inductive}~\textit{even}~\textbf{where} \\
``\textit{even}~0'' $\,\mid$ \\
``\textit{even}~$n\,\Longrightarrow\, \textit{even}~(\textit{Suc}~(\textit{Suc}~n))$''
\postw

This predicate enjoys the desirable property of being well-founded, which means
that the introduction rules don't give rise to infinite chains of the form

\prew
$\cdots\,\Longrightarrow\, \textit{even}~k''
       \,\Longrightarrow\, \textit{even}~k'
       \,\Longrightarrow\, \textit{even}~k.$
\postw

For \textit{even}, this is obvious: Any chain ending at $k$ will be of length
$k/2 + 1$:

\prew
$\textit{even}~0\,\Longrightarrow\, \textit{even}~2\,\Longrightarrow\, \cdots
       \,\Longrightarrow\, \textit{even}~(k - 2)
       \,\Longrightarrow\, \textit{even}~k.$
\postw

Wellfoundedness is desirable because it enables Nitpick to use a very efficient
fixed point computation.%
\footnote{If an inductive predicate is
well-founded, then it has exactly one fixed point, which is simultaneously the
least and the greatest fixed point. In these circumstances, the computation of
the least fixed point amounts to the computation of an arbitrary fixed point,
which can be performed using a straightforward recursive equation.}
Moreover, Nitpick can prove wellfoundedness of most well-founded predicates,
just as Isabelle's \textbf{function} package usually discharges termination
proof obligations automatically.

Let's try an example:

\prew
\textbf{lemma} ``$\exists n.\; \textit{even}~n \mathrel{\land} \textit{even}~(\textit{Suc}~n)$'' \\
\textbf{nitpick}~[\textit{card nat}~= 100, \textit{unary\_ints}, \textit{verbose}] \\[2\smallskipamount]
\slshape The inductive predicate ``\textit{even}'' was proved well-founded.
Nitpick can compute it efficiently. \\[2\smallskipamount]
Trying 1 scope: \\
\hbox{}\qquad \textit{card nat}~= 100. \\[2\smallskipamount]
Nitpick found a potential counterexample for \textit{card nat}~= 100: \\[2\smallskipamount]
\hbox{}\qquad Empty assignment \\[2\smallskipamount]
Nitpick could not find a better counterexample. \\[2\smallskipamount]
Total time: 2274 ms.
\postw

No genuine counterexample is possible because Nitpick cannot rule out the
existence of a natural number $n \ge 100$ such that both $\textit{even}~n$ and
$\textit{even}~(\textit{Suc}~n)$ are true. To help Nitpick, we can bound the
existential quantifier:

\prew
\textbf{lemma} ``$\exists n \mathbin{\le} 99.\; \textit{even}~n \mathrel{\land} \textit{even}~(\textit{Suc}~n)$'' \\
\textbf{nitpick}~[\textit{card nat}~= 100, \textit{unary\_ints}] \\[2\smallskipamount]
\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Empty assignment
\postw

So far we were blessed by the wellfoundedness of \textit{even}. What happens if
we use the following definition instead?

\prew
\textbf{inductive} $\textit{even}'$ \textbf{where} \\
``$\textit{even}'~(0{\Colon}\textit{nat})$'' $\,\mid$ \\
``$\textit{even}'~2$'' $\,\mid$ \\
``$\lbrakk\textit{even}'~m;\> \textit{even}'~n\rbrakk \,\Longrightarrow\, \textit{even}'~(m + n)$''
\postw

This definition is not well-founded: From $\textit{even}'~0$ and
$\textit{even}'~0$, we can derive that $\textit{even}'~0$. Nonetheless, the
predicates $\textit{even}$ and $\textit{even}'$ are equivalent.

Let's check a property involving $\textit{even}'$. To make up for the
foreseeable computational hurdles entailed by non-wellfoundedness, we decrease
\textit{nat}'s cardinality to a mere 10:

\prew
\textbf{lemma}~``$\exists n \in \{0, 2, 4, 6, 8\}.\;
\lnot\;\textit{even}'~n$'' \\
\textbf{nitpick}~[\textit{card nat}~= 10,\, \textit{verbose},\, \textit{show\_consts}] \\[2\smallskipamount]
\slshape
The inductive predicate ``$\textit{even}'\!$'' could not be proved well-founded.
Nitpick might need to unroll it. \\[2\smallskipamount]
Trying 6 scopes: \\
\hbox{}\qquad \textit{card nat}~= 10 and \textit{iter} $\textit{even}'$~= 0; \\
\hbox{}\qquad \textit{card nat}~= 10 and \textit{iter} $\textit{even}'$~= 1; \\
\hbox{}\qquad \textit{card nat}~= 10 and \textit{iter} $\textit{even}'$~= 2; \\
\hbox{}\qquad \textit{card nat}~= 10 and \textit{iter} $\textit{even}'$~= 4; \\
\hbox{}\qquad \textit{card nat}~= 10 and \textit{iter} $\textit{even}'$~= 8; \\
\hbox{}\qquad \textit{card nat}~= 10 and \textit{iter} $\textit{even}'$~= 9. \\[2\smallskipamount]
Nitpick found a counterexample for \textit{card nat}~= 10 and \textit{iter} $\textit{even}'$~= 2: \\[2\smallskipamount]
\hbox{}\qquad Constant: \nopagebreak \\
\hbox{}\qquad\qquad $\lambda i.\; \textit{even}'$ = $\undef(\!\begin{aligned}[t]
& 2 := \{0, 2, 4, 6, 8, 1^\Q, 3^\Q, 5^\Q, 7^\Q, 9^\Q\}, \\[-2pt]
& 1 := \{0, 2, 4, 1^\Q, 3^\Q, 5^\Q, 6^\Q, 7^\Q, 8^\Q, 9^\Q\}, \\[-2pt]
& 0 := \{0, 2, 1^\Q, 3^\Q, 4^\Q, 5^\Q, 6^\Q, 7^\Q, 8^\Q, 9^\Q\})\end{aligned}$ \\[2\smallskipamount]
Total time: 1140 ms.
\postw

Nitpick's output is very instructive. First, it tells us that the predicate is
unrolled, meaning that it is computed iteratively from the empty set. Then it
lists six scopes specifying different bounds on the numbers of iterations:\ 0,
1, 2, 4, 8, and~9.

The output also shows how each iteration contributes to $\textit{even}'$. The
notation $\lambda i.\; \textit{even}'$ indicates that the value of the
predicate depends on an iteration counter. Iteration 0 provides the basis
elements, $0$ and $2$. Iteration 1 contributes $4$ ($= 2 + 2$). Iteration 2
throws $6$ ($= 2 + 4 = 4 + 2$) and $8$ ($= 4 + 4$) into the mix. Further
iterations would not contribute any new elements.

Some values are marked with superscripted question
marks~(`\lower.2ex\hbox{$^\Q$}'). These are the elements for which the
predicate evaluates to $\unk$. Thus, $\textit{even}'$ evaluates to either
\textit{True} or $\unk$, never \textit{False}.

When unrolling a predicate, Nitpick tries 0, 1, 2, 4, 8, 12, 16, and 24
iterations. However, these numbers are bounded by the cardinality of the
predicate's domain. With \textit{card~nat}~= 10, no more than 9 iterations are
ever needed to compute the value of a \textit{nat} predicate. You can specify
the number of iterations using the \textit{iter} option, as explained in
\S\ref{scope-of-search}.

In the next formula, $\textit{even}'$ occurs both positively and negatively:

\prew
\textbf{lemma} ``$\textit{even}'~(n - 2) \,\Longrightarrow\, \textit{even}'~n$'' \\
\textbf{nitpick} [\textit{card nat} = 10, \textit{show\_consts}] \\[2\smallskipamount]
\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Free variable: \nopagebreak \\
\hbox{}\qquad\qquad $n = 1$ \\
\hbox{}\qquad Constants: \nopagebreak \\
\hbox{}\qquad\qquad $\lambda i.\; \textit{even}'$ = $\undef(\!\begin{aligned}[t]
& 0 := \{0, 2, 1^\Q, 3^\Q, 4^\Q, 5^\Q, 6^\Q, 7^\Q, 8^\Q, 9^\Q\})\end{aligned}$  \\
\hbox{}\qquad\qquad $\textit{even}' \subseteq \{0, 2, 4, 6, 8, \unr\}$
\postw

Notice the special constraint $\textit{even}' \subseteq \{0,\, 2,\, 4,\, 6,\,
8,\, \unr\}$ in the output, whose right-hand side represents an arbitrary
fixed point (not necessarily the least one). It is used to falsify
$\textit{even}'~n$. In contrast, the unrolled predicate is used to satisfy
$\textit{even}'~(n - 2)$.

Coinductive predicates are handled dually. For example:

\prew
\textbf{coinductive} \textit{nats} \textbf{where} \\
``$\textit{nats}~(x\Colon\textit{nat}) \,\Longrightarrow\, \textit{nats}~x$'' \\[2\smallskipamount]
\textbf{lemma} ``$\textit{nats} = \{0, 1, 2, 3, 4\}$'' \\
\textbf{nitpick}~[\textit{card nat} = 10,\, \textit{show\_consts}] \\[2\smallskipamount]
\slshape Nitpick found a counterexample:
\\[2\smallskipamount]
\hbox{}\qquad Constants: \nopagebreak \\
\hbox{}\qquad\qquad $\lambda i.\; \textit{nats} = \undef(0 := \{\!\begin{aligned}[t]
& 0^\Q, 1^\Q, 2^\Q, 3^\Q, 4^\Q, 5^\Q, 6^\Q, 7^\Q, 8^\Q, 9^\Q, \\[-2pt]
& \unr\})\end{aligned}$ \\
\hbox{}\qquad\qquad $nats \supseteq \{9, 5^\Q, 6^\Q, 7^\Q, 8^\Q, \unr\}$
\postw

As a special case, Nitpick uses Kodkod's transitive closure operator to encode
negative occurrences of non-well-founded ``linear inductive predicates,'' i.e.,
inductive predicates for which each the predicate occurs in at most one
assumption of each introduction rule. For example:

\prew
\textbf{inductive} \textit{odd} \textbf{where} \\
``$\textit{odd}~1$'' $\,\mid$ \\
``$\lbrakk \textit{odd}~m;\>\, \textit{even}~n\rbrakk \,\Longrightarrow\, \textit{odd}~(m + n)$'' \\[2\smallskipamount]
\textbf{lemma}~``$\textit{odd}~n \,\Longrightarrow\, \textit{odd}~(n - 2)$'' \\
\textbf{nitpick}~[\textit{card nat} = 10,\, \textit{show\_consts}] \\[2\smallskipamount]
\slshape Nitpick found a counterexample:
\\[2\smallskipamount]
\hbox{}\qquad Free variable: \nopagebreak \\
\hbox{}\qquad\qquad $n = 1$ \\
\hbox{}\qquad Constants: \nopagebreak \\
\hbox{}\qquad\qquad $\textit{even} = \{0, 2, 4, 6, 8, \unr\}$ \\
\hbox{}\qquad\qquad $\textit{odd}_{\textsl{base}} = \{1, \unr\}$ \\
\hbox{}\qquad\qquad $\textit{odd}_{\textsl{step}} = \!
\!\begin{aligned}[t]
  & \{(0, 0), (0, 2), (0, 4), (0, 6), (0, 8), (1, 1), (1, 3), (1, 5), \\[-2pt]
  & \phantom{\{} (1, 7), (1, 9), (2, 2), (2, 4), (2, 6), (2, 8), (3, 3),
       (3, 5), \\[-2pt]
  & \phantom{\{} (3, 7), (3, 9), (4, 4), (4, 6), (4, 8), (5, 5), (5, 7), (5, 9), \\[-2pt]
  & \phantom{\{} (6, 6), (6, 8), (7, 7), (7, 9), (8, 8), (9, 9), \unr\}\end{aligned}$ \\
\hbox{}\qquad\qquad $\textit{odd} \subseteq \{1, 3, 5, 7, 9, 8^\Q, \unr\}$
\postw

\noindent
In the output, $\textit{odd}_{\textrm{base}}$ represents the base elements and
$\textit{odd}_{\textrm{step}}$ is a transition relation that computes new
elements from known ones. The set $\textit{odd}$ consists of all the values
reachable through the reflexive transitive closure of
$\textit{odd}_{\textrm{step}}$ starting with any element from
$\textit{odd}_{\textrm{base}}$, namely 1, 3, 5, 7, and 9. Using Kodkod's
transitive closure to encode linear predicates is normally either more thorough
or more efficient than unrolling (depending on the value of \textit{iter}), but
for those cases where it isn't you can disable it by passing the
\textit{dont\_star\_linear\_preds} option.

\subsection{Coinductive Datatypes}
\label{coinductive-datatypes}

While Isabelle regrettably lacks a high-level mechanism for defining coinductive
datatypes, the \textit{Coinductive\_List} theory provides a coinductive ``lazy
list'' datatype, $'a~\textit{llist}$, defined the hard way. Nitpick supports
these lazy lists seamlessly and provides a hook, described in
\S\ref{registration-of-coinductive-datatypes}, to register custom coinductive
datatypes.

(Co)intuitively, a coinductive datatype is similar to an inductive datatype but
allows infinite objects. Thus, the infinite lists $\textit{ps}$ $=$ $[a, a, a,
\ldots]$, $\textit{qs}$ $=$ $[a, b, a, b, \ldots]$, and $\textit{rs}$ $=$ $[0,
1, 2, 3, \ldots]$ can be defined as lazy lists using the
$\textit{LNil}\mathbin{\Colon}{'}a~\textit{llist}$ and
$\textit{LCons}\mathbin{\Colon}{'}a \mathbin{\Rightarrow} {'}a~\textit{llist}
\mathbin{\Rightarrow} {'}a~\textit{llist}$ constructors.

Although it is otherwise no friend of infinity, Nitpick can find counterexamples
involving cyclic lists such as \textit{ps} and \textit{qs} above as well as
finite lists:

\prew
\textbf{lemma} ``$\textit{xs} \not= \textit{LCons}~a~\textit{xs}$'' \\
\textbf{nitpick} \\[2\smallskipamount]
\slshape Nitpick found a counterexample for {\itshape card}~$'a$ = 1: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $\textit{a} = a_1$ \\
\hbox{}\qquad\qquad $\textit{xs} = \textsl{THE}~\omega.\; \omega = \textit{LCons}~a_1~\omega$
\postw

The notation $\textrm{THE}~\omega.\; \omega = t(\omega)$ stands
for the infinite term $t(t(t(\ldots)))$. Hence, \textit{xs} is simply the
infinite list $[a_1, a_1, a_1, \ldots]$.

The next example is more interesting:

\prew
\textbf{lemma}~``$\lbrakk\textit{xs} = \textit{LCons}~a~\textit{xs};\>\,
\textit{ys} = \textit{iterates}~(\lambda b.\> a)~b\rbrakk \,\Longrightarrow\, \textit{xs} = \textit{ys}$'' \\
\textbf{nitpick} [\textit{verbose}] \\[2\smallskipamount]
\slshape The type ``\kern1pt$'a$'' passed the monotonicity test. Nitpick might be able to skip
some scopes. \\[2\smallskipamount]
Trying 8 scopes: \\
\hbox{}\qquad \textit{card} $'a$~= 1, \textit{card} ``\kern1pt$'a~\textit{list}$''~= 1,
and \textit{bisim\_depth}~= 0. \\
\hbox{}\qquad $\qquad\vdots$ \\[.5\smallskipamount]
\hbox{}\qquad \textit{card} $'a$~= 8, \textit{card} ``\kern1pt$'a~\textit{list}$''~= 8,
and \textit{bisim\_depth}~= 7. \\[2\smallskipamount]
Nitpick found a counterexample for {\itshape card}~$'a$ = 2,
\textit{card}~``\kern1pt$'a~\textit{list}$''~= 2, and \textit{bisim\_\allowbreak
depth}~= 1:
\\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $\textit{a} = a_2$ \\
\hbox{}\qquad\qquad $\textit{b} = a_1$ \\
\hbox{}\qquad\qquad $\textit{xs} = \textsl{THE}~\omega.\; \omega = \textit{LCons}~a_2~\omega$ \\
\hbox{}\qquad\qquad $\textit{ys} = \textit{LCons}~a_1~(\textsl{THE}~\omega.\; \omega = \textit{LCons}~a_2~\omega)$ \\[2\smallskipamount]
Total time: 726 ms.
\postw

The lazy list $\textit{xs}$ is simply $[a_2, a_2, a_2, \ldots]$, whereas
$\textit{ys}$ is $[a_1, a_2, a_2, a_2, \ldots]$, i.e., a lasso-shaped list with
$[a_1]$ as its stem and $[a_2]$ as its cycle. In general, the list segment
within the scope of the {THE} binder corresponds to the lasso's cycle, whereas
the segment leading to the binder is the stem.

A salient property of coinductive datatypes is that two objects are considered
equal if and only if they lead to the same observations. For example, the lazy
lists $\textrm{THE}~\omega.\; \omega =
\textit{LCons}~a~(\textit{LCons}~b~\omega)$ and
$\textit{LCons}~a~(\textrm{THE}~\omega.\; \omega =
\textit{LCons}~b~(\textit{LCons}~a~\omega))$ are identical, because both lead
to the sequence of observations $a$, $b$, $a$, $b$, \hbox{\ldots} (or,
equivalently, both encode the infinite list $[a, b, a, b, \ldots]$). This
concept of equality for coinductive datatypes is called bisimulation and is
defined coinductively.

Internally, Nitpick encodes the coinductive bisimilarity predicate as part of
the Kodkod problem to ensure that distinct objects lead to different
observations. This precaution is somewhat expensive and often unnecessary, so it
can be disabled by setting the \textit{bisim\_depth} option to $-1$. The
bisimilarity check is then performed \textsl{after} the counterexample has been
found to ensure correctness. If this after-the-fact check fails, the
counterexample is tagged as ``likely genuine'' and Nitpick recommends to try
again with \textit{bisim\_depth} set to a nonnegative integer. Disabling the
check for the previous example saves approximately 150~milli\-seconds; the speed
gains can be more significant for larger scopes.

The next formula illustrates the need for bisimilarity (either as a Kodkod
predicate or as an after-the-fact check) to prevent spurious counterexamples:

\prew
\textbf{lemma} ``$\lbrakk xs = \textit{LCons}~a~\textit{xs};\>\, \textit{ys} = \textit{LCons}~a~\textit{ys}\rbrakk
\,\Longrightarrow\, \textit{xs} = \textit{ys}$'' \\
\textbf{nitpick} [\textit{bisim\_depth} = $-1$, \textit{show\_datatypes}] \\[2\smallskipamount]
\slshape Nitpick found a likely genuine counterexample for $\textit{card}~'a$ = 2: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $a = a_2$ \\
\hbox{}\qquad\qquad $\textit{xs} = \textsl{THE}~\omega.\; \omega =
\textit{LCons}~a_2~\omega$ \\
\hbox{}\qquad\qquad $\textit{ys} = \textsl{THE}~\omega.\; \omega = \textit{LCons}~a_2~\omega$ \\
\hbox{}\qquad Codatatype:\strut \nopagebreak \\
\hbox{}\qquad\qquad $'a~\textit{llist} =
\{\!\begin{aligned}[t]
  & \textsl{THE}~\omega.\; \omega = \textit{LCons}~a_2~\omega, \\[-2pt]
  & \textsl{THE}~\omega.\; \omega = \textit{LCons}~a_2~\omega,\> \unr\}\end{aligned}$
\\[2\smallskipamount]
Try again with ``\textit{bisim\_depth}'' set to a nonnegative value to confirm
that the counterexample is genuine. \\[2\smallskipamount]
{\upshape\textbf{nitpick}} \\[2\smallskipamount]
\slshape Nitpick found no counterexample.
\postw

In the first \textbf{nitpick} invocation, the after-the-fact check discovered 
that the two known elements of type $'a~\textit{llist}$ are bisimilar.

A compromise between leaving out the bisimilarity predicate from the Kodkod
problem and performing the after-the-fact check is to specify a lower
nonnegative \textit{bisim\_depth} value than the default one provided by
Nitpick. In general, a value of $K$ means that Nitpick will require all lists to
be distinguished from each other by their prefixes of length $K$. Be aware that
setting $K$ to a too low value can overconstrain Nitpick, preventing it from
finding any counterexamples.

\subsection{Boxing}
\label{boxing}

Nitpick normally maps function and product types directly to the corresponding
Kodkod concepts. As a consequence, if $'a$ has cardinality 3 and $'b$ has
cardinality 4, then $'a \times {'}b$ has cardinality 12 ($= 4 \times 3$) and $'a
\Rightarrow {'}b$ has cardinality 64 ($= 4^3$). In some circumstances, it pays
off to treat these types in the same way as plain datatypes, by approximating
them by a subset of a given cardinality. This technique is called ``boxing'' and
is particularly useful for functions passed as arguments to other functions, for
high-arity functions, and for large tuples. Under the hood, boxing involves
wrapping occurrences of the types $'a \times {'}b$ and $'a \Rightarrow {'}b$ in
isomorphic datatypes, as can be seen by enabling the \textit{debug} option.

To illustrate boxing, we consider a formalization of $\lambda$-terms represented
using de Bruijn's notation:

\prew
\textbf{datatype} \textit{tm} = \textit{Var}~\textit{nat}~$\mid$~\textit{Lam}~\textit{tm} $\mid$ \textit{App~tm~tm}
\postw

The $\textit{lift}~t~k$ function increments all variables with indices greater
than or equal to $k$ by one:

\prew
\textbf{primrec} \textit{lift} \textbf{where} \\
``$\textit{lift}~(\textit{Var}~j)~k = \textit{Var}~(\textrm{if}~j < k~\textrm{then}~j~\textrm{else}~j + 1)$'' $\mid$ \\
``$\textit{lift}~(\textit{Lam}~t)~k = \textit{Lam}~(\textit{lift}~t~(k + 1))$'' $\mid$ \\
``$\textit{lift}~(\textit{App}~t~u)~k = \textit{App}~(\textit{lift}~t~k)~(\textit{lift}~u~k)$''
\postw

The $\textit{loose}~t~k$ predicate returns \textit{True} if and only if
term $t$ has a loose variable with index $k$ or more:

\prew
\textbf{primrec}~\textit{loose} \textbf{where} \\
``$\textit{loose}~(\textit{Var}~j)~k = (j \ge k)$'' $\mid$ \\
``$\textit{loose}~(\textit{Lam}~t)~k = \textit{loose}~t~(\textit{Suc}~k)$'' $\mid$ \\
``$\textit{loose}~(\textit{App}~t~u)~k = (\textit{loose}~t~k \mathrel{\lor} \textit{loose}~u~k)$''
\postw

Next, the $\textit{subst}~\sigma~t$ function applies the substitution $\sigma$
on $t$:

\prew
\textbf{primrec}~\textit{subst} \textbf{where} \\
``$\textit{subst}~\sigma~(\textit{Var}~j) = \sigma~j$'' $\mid$ \\
``$\textit{subst}~\sigma~(\textit{Lam}~t) = {}$\phantom{''} \\
\phantom{``}$\textit{Lam}~(\textit{subst}~(\lambda n.\> \textrm{case}~n~\textrm{of}~0 \Rightarrow \textit{Var}~0 \mid \textit{Suc}~m \Rightarrow \textit{lift}~(\sigma~m)~1)~t)$'' $\mid$ \\
``$\textit{subst}~\sigma~(\textit{App}~t~u) = \textit{App}~(\textit{subst}~\sigma~t)~(\textit{subst}~\sigma~u)$''
\postw

A substitution is a function that maps variable indices to terms. Observe that
$\sigma$ is a function passed as argument and that Nitpick can't optimize it
away, because the recursive call for the \textit{Lam} case involves an altered
version. Also notice the \textit{lift} call, which increments the variable
indices when moving under a \textit{Lam}.

A reasonable property to expect of substitution is that it should leave closed
terms unchanged. Alas, even this simple property does not hold:

\pre
\textbf{lemma}~``$\lnot\,\textit{loose}~t~0 \,\Longrightarrow\, \textit{subst}~\sigma~t = t$'' \\
\textbf{nitpick} [\textit{verbose}] \\[2\smallskipamount]
\slshape
Trying 8 scopes: \nopagebreak \\
\hbox{}\qquad \textit{card~nat}~= 1, \textit{card tm}~= 1, and \textit{card} ``$\textit{nat} \Rightarrow \textit{tm}$'' = 1; \\
\hbox{}\qquad \textit{card~nat}~= 2, \textit{card tm}~= 2, and \textit{card} ``$\textit{nat} \Rightarrow \textit{tm}$'' = 2; \\
\hbox{}\qquad $\qquad\vdots$ \\[.5\smallskipamount]
\hbox{}\qquad \textit{card~nat}~= 8, \textit{card tm}~= 8, and \textit{card} ``$\textit{nat} \Rightarrow \textit{tm}$'' = 8. \\[2\smallskipamount]
Nitpick found a counterexample for \textit{card~nat}~= 6, \textit{card~tm}~= 6,
and \textit{card}~``$\textit{nat} \Rightarrow \textit{tm}$''~= 6: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $\sigma = \undef(\!\begin{aligned}[t]
& 0 := \textit{Var}~0,\>
  1 := \textit{Var}~0,\>
  2 := \textit{Var}~0, \\[-2pt]
& 3 := \textit{Var}~0,\>
  4 := \textit{Var}~0,\>
  5 := \textit{Var}~0)\end{aligned}$ \\
\hbox{}\qquad\qquad $t = \textit{Lam}~(\textit{Lam}~(\textit{Var}~1))$ \\[2\smallskipamount]
Total time: $4679$ ms.
\postw

Using \textit{eval}, we find out that $\textit{subst}~\sigma~t =
\textit{Lam}~(\textit{Lam}~(\textit{Var}~0))$. Using the traditional
$\lambda$-term notation, $t$~is
$\lambda x\, y.\> x$ whereas $\textit{subst}~\sigma~t$ is $\lambda x\, y.\> y$.
The bug is in \textit{subst}: The $\textit{lift}~(\sigma~m)~1$ call should be
replaced with $\textit{lift}~(\sigma~m)~0$.

An interesting aspect of Nitpick's verbose output is that it assigned inceasing
cardinalities from 1 to 8 to the type $\textit{nat} \Rightarrow \textit{tm}$.
For the formula of interest, knowing 6 values of that type was enough to find
the counterexample. Without boxing, $46\,656$ ($= 6^6$) values must be
considered, a hopeless undertaking:

\prew
\textbf{nitpick} [\textit{dont\_box}] \\[2\smallskipamount]
{\slshape Nitpick ran out of time after checking 4 of 8 scopes.}
\postw

{\looseness=-1
Boxing can be enabled or disabled globally or on a per-type basis using the
\textit{box} option. Moreover, setting the cardinality of a function or
product type implicitly enables boxing for that type. Nitpick usually performs
reasonable choices about which types should be boxed, but option tweaking
sometimes helps.

}

\subsection{Scope Monotonicity}
\label{scope-monotonicity}

The \textit{card} option (together with \textit{iter}, \textit{bisim\_depth},
and \textit{max}) controls which scopes are actually tested. In general, to
exhaust all models below a certain cardinality bound, the number of scopes that
Nitpick must consider increases exponentially with the number of type variables
(and \textbf{typedecl}'d types) occurring in the formula. Given the default
cardinality specification of 1--8, no fewer than $8^4 = 4096$ scopes must be
considered for a formula involving $'a$, $'b$, $'c$, and $'d$.

Fortunately, many formulas exhibit a property called \textsl{scope
monotonicity}, meaning that if the formula is falsifiable for a given scope,
it is also falsifiable for all larger scopes \cite[p.~165]{jackson-2006}.

Consider the formula

\prew
\textbf{lemma}~``$\textit{length~xs} = \textit{length~ys} \,\Longrightarrow\, \textit{rev}~(\textit{zip~xs~ys}) = \textit{zip~xs}~(\textit{rev~ys})$''
\postw

where \textit{xs} is of type $'a~\textit{list}$ and \textit{ys} is of type
$'b~\textit{list}$. A priori, Nitpick would need to consider 512 scopes to
exhaust the specification \textit{card}~= 1--8. However, our intuition tells us
that any counterexample found with a small scope would still be a counterexample
in a larger scope---by simply ignoring the fresh $'a$ and $'b$ values provided
by the larger scope. Nitpick comes to the same conclusion after a careful
inspection of the formula and the relevant definitions:

\prew
\textbf{nitpick}~[\textit{verbose}] \\[2\smallskipamount]
\slshape
The types ``\kern1pt$'a$'' and ``\kern1pt$'b$'' passed the monotonicity test.
Nitpick might be able to skip some scopes.
 \\[2\smallskipamount]
Trying 8 scopes: \\
\hbox{}\qquad \textit{card} $'a$~= 1, \textit{card} $'b$~= 1,
\textit{card} \textit{nat}~= 1, \textit{card} ``$('a \times {'}b)$
\textit{list}''~= 1, \\
\hbox{}\qquad\quad \textit{card} ``\kern1pt$'a$ \textit{list}''~= 1, and
\textit{card} ``\kern1pt$'b$ \textit{list}''~= 1. \\
\hbox{}\qquad \textit{card} $'a$~= 2, \textit{card} $'b$~= 2,
\textit{card} \textit{nat}~= 2, \textit{card} ``$('a \times {'}b)$
\textit{list}''~= 2, \\
\hbox{}\qquad\quad \textit{card} ``\kern1pt$'a$ \textit{list}''~= 2, and
\textit{card} ``\kern1pt$'b$ \textit{list}''~= 2. \\
\hbox{}\qquad $\qquad\vdots$ \\[.5\smallskipamount]
\hbox{}\qquad \textit{card} $'a$~= 8, \textit{card} $'b$~= 8,
\textit{card} \textit{nat}~= 8, \textit{card} ``$('a \times {'}b)$
\textit{list}''~= 8, \\
\hbox{}\qquad\quad \textit{card} ``\kern1pt$'a$ \textit{list}''~= 8, and
\textit{card} ``\kern1pt$'b$ \textit{list}''~= 8.
\\[2\smallskipamount]
Nitpick found a counterexample for
\textit{card} $'a$~= 5, \textit{card} $'b$~= 5,
\textit{card} \textit{nat}~= 5, \textit{card} ``$('a \times {'}b)$
\textit{list}''~= 5, \textit{card} ``\kern1pt$'a$ \textit{list}''~= 5, and
\textit{card} ``\kern1pt$'b$ \textit{list}''~= 5:
\\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $\textit{xs} = [a_4, a_5]$ \\
\hbox{}\qquad\qquad $\textit{ys} = [b_3, b_3]$ \\[2\smallskipamount]
Total time: 1636 ms.
\postw

In theory, it should be sufficient to test a single scope:

\prew
\textbf{nitpick}~[\textit{card}~= 8]
\postw

However, this is often less efficient in practice and may lead to overly complex
counterexamples.

If the monotonicity check fails but we believe that the formula is monotonic (or
we don't mind missing some counterexamples), we can pass the
\textit{mono} option. To convince yourself that this option is risky,
simply consider this example from \S\ref{skolemization}:

\prew
\textbf{lemma} ``$\exists g.\; \forall x\Colon 'b.~g~(f~x) = x
 \,\Longrightarrow\, \forall y\Colon {'}a.\; \exists x.~y = f~x$'' \\
\textbf{nitpick} [\textit{mono}] \\[2\smallskipamount]
{\slshape Nitpick found no counterexample.} \\[2\smallskipamount]
\textbf{nitpick} \\[2\smallskipamount]
\slshape
Nitpick found a counterexample for \textit{card} $'a$~= 2 and \textit{card} $'b$~=~1: \\
\hbox{}\qquad $\vdots$
\postw

(It turns out the formula holds if and only if $\textit{card}~'a \le
\textit{card}~'b$.) Although this is rarely advisable, the automatic
monotonicity checks can be disabled by passing \textit{non\_mono}
(\S\ref{optimizations}).

As insinuated in \S\ref{natural-numbers-and-integers} and
\S\ref{inductive-datatypes}, \textit{nat}, \textit{int}, and inductive datatypes
are normally monotonic and treated as such. The same is true for record types,
\textit{rat}, \textit{real}, and some \textbf{typedef}'d types. Thus, given the
cardinality specification 1--8, a formula involving \textit{nat}, \textit{int},
\textit{int~list}, \textit{rat}, and \textit{rat~list} will lead Nitpick to
consider only 8~scopes instead of $32\,768$.

\subsection{Inductive Properties}
\label{inductive-properties}

Inductive properties are a particular pain to prove, because the failure to
establish an induction step can mean several things:
%
\begin{enumerate}
\item The property is invalid.
\item The property is valid but is too weak to support the induction step.
\item The property is valid and strong enough; it's just that we haven't found
the proof yet.
\end{enumerate}
%
Depending on which scenario applies, we would take the appropriate course of
action:
%
\begin{enumerate}
\item Repair the statement of the property so that it becomes valid.
\item Generalize the property and/or prove auxiliary properties.
\item Work harder on a proof.
\end{enumerate}
%
How can we distinguish between the three scenarios? Nitpick's normal mode of
operation can often detect scenario 1, and Isabelle's automatic tactics help with
scenario 3. Using appropriate techniques, it is also often possible to use
Nitpick to identify scenario 2. Consider the following transition system,
in which natural numbers represent states:

\prew
\textbf{inductive\_set}~\textit{reach}~\textbf{where} \\
``$(4\Colon\textit{nat}) \in \textit{reach\/}$'' $\mid$ \\
``$\lbrakk n < 4;\> n \in \textit{reach\/}\rbrakk \,\Longrightarrow\, 3 * n + 1 \in \textit{reach\/}$'' $\mid$ \\
``$n \in \textit{reach} \,\Longrightarrow n + 2 \in \textit{reach\/}$''
\postw

We will try to prove that only even numbers are reachable:

\prew
\textbf{lemma}~``$n \in \textit{reach} \,\Longrightarrow\, 2~\textrm{dvd}~n$''
\postw

Does this property hold? Nitpick cannot find a counterexample within 30 seconds,
so let's attempt a proof by induction:

\prew
\textbf{apply}~(\textit{induct~set}{:}~\textit{reach\/}) \\
\textbf{apply}~\textit{auto}
\postw

This leaves us in the following proof state:

\prew
{\slshape goal (2 subgoals): \\
\phantom{0}1. ${\bigwedge}n.\;\, \lbrakk n \in \textit{reach\/};\, n < 4;\, 2~\textsl{dvd}~n\rbrakk \,\Longrightarrow\, 2~\textsl{dvd}~\textit{Suc}~(3 * n)$ \\
\phantom{0}2. ${\bigwedge}n.\;\, \lbrakk n \in \textit{reach\/};\, 2~\textsl{dvd}~n\rbrakk \,\Longrightarrow\, 2~\textsl{dvd}~\textit{Suc}~(\textit{Suc}~n)$
}
\postw

If we run Nitpick on the first subgoal, it still won't find any
counterexample; and yet, \textit{auto} fails to go further, and \textit{arith}
is helpless. However, notice the $n \in \textit{reach}$ assumption, which
strengthens the induction hypothesis but is not immediately usable in the proof.
If we remove it and invoke Nitpick, this time we get a counterexample:

\prew
\textbf{apply}~(\textit{thin\_tac}~``$n \in \textit{reach\/}$'') \\
\textbf{nitpick} \\[2\smallskipamount]
\slshape Nitpick found a counterexample: \\[2\smallskipamount]
\hbox{}\qquad Skolem constant: \nopagebreak \\
\hbox{}\qquad\qquad $n = 0$
\postw

Indeed, 0 < 4, 2 divides 0, but 2 does not divide 1. We can use this information
to strength the lemma:

\prew
\textbf{lemma}~``$n \in \textit{reach} \,\Longrightarrow\, 2~\textrm{dvd}~n \mathrel{\lor} n \not= 0$''
\postw

Unfortunately, the proof by induction still gets stuck, except that Nitpick now
finds the counterexample $n = 2$. We generalize the lemma further to

\prew
\textbf{lemma}~``$n \in \textit{reach} \,\Longrightarrow\, 2~\textrm{dvd}~n \mathrel{\lor} n \ge 4$''
\postw

and this time \textit{arith} can finish off the subgoals.

A similar technique can be employed for structural induction. The
following mini-formalization of full binary trees will serve as illustration:

\prew
\textbf{datatype} $\kern1pt'a$~\textit{bin\_tree} = $\textit{Leaf}~{\kern1pt'a}$ $\mid$ $\textit{Branch}$ ``\kern1pt$'a$ \textit{bin\_tree}'' ``\kern1pt$'a$ \textit{bin\_tree}'' \\[2\smallskipamount]
\textbf{primrec}~\textit{labels}~\textbf{where} \\
``$\textit{labels}~(\textit{Leaf}~a) = \{a\}$'' $\mid$ \\
``$\textit{labels}~(\textit{Branch}~t~u) = \textit{labels}~t \mathrel{\cup} \textit{labels}~u$'' \\[2\smallskipamount]
\textbf{primrec}~\textit{swap}~\textbf{where} \\
``$\textit{swap}~(\textit{Leaf}~c)~a~b =$ \\
\phantom{``}$(\textrm{if}~c = a~\textrm{then}~\textit{Leaf}~b~\textrm{else~if}~c = b~\textrm{then}~\textit{Leaf}~a~\textrm{else}~\textit{Leaf}~c)$'' $\mid$ \\
``$\textit{swap}~(\textit{Branch}~t~u)~a~b = \textit{Branch}~(\textit{swap}~t~a~b)~(\textit{swap}~u~a~b)$''
\postw

The \textit{labels} function returns the set of labels occurring on leaves of a
tree, and \textit{swap} exchanges two labels. Intuitively, if two distinct
labels $a$ and $b$ occur in a tree $t$, they should also occur in the tree
obtained by swapping $a$ and $b$:

\prew
\textbf{lemma} $``\lbrakk a \in \textit{labels}~t;\, b \in \textit{labels}~t;\, a \not= b\rbrakk {}$ \\
\phantom{\textbf{lemma} ``}$\,{\Longrightarrow}{\;\,} \textit{labels}~(\textit{swap}~t~a~b) = \textit{labels}~t$''
\postw

Nitpick can't find any counterexample, so we proceed with induction
(this time favoring a more structured style):

\prew
\textbf{proof}~(\textit{induct}~$t$) \\
\hbox{}\quad \textbf{case}~\textit{Leaf}~\textbf{thus}~\textit{?case}~\textbf{by}~\textit{simp} \\
\textbf{next} \\
\hbox{}\quad \textbf{case}~$(\textit{Branch}~t~u)$~\textbf{thus} \textit{?case}
\postw

Nitpick can't find any counterexample at this point either, but it makes the
following suggestion:

\prew
\slshape
Hint: To check that the induction hypothesis is general enough, try the following command:
\textbf{nitpick}~[\textit{non\_std} ``${\kern1pt'a}~\textit{bin\_tree}$'', \textit{show\_consts}].
\postw

If we follow the hint, we get a ``nonstandard'' counterexample for the step:

\prew
\slshape Nitpick found a nonstandard counterexample for \textit{card} $'a$ = 4: \\[2\smallskipamount]
\hbox{}\qquad Free variables: \nopagebreak \\
\hbox{}\qquad\qquad $a = a_4$ \\
\hbox{}\qquad\qquad $b = a_3$ \\
\hbox{}\qquad\qquad $t = \xi_3$ \\
\hbox{}\qquad\qquad $u = \xi_4$ \\
\hbox{}\qquad {\slshape Constants:} \nopagebreak \\
\hbox{}\qquad\qquad $\textit{labels} = \undef
    (\!\begin{aligned}[t]%
    & \xi_3 := \{a_4\},\> \xi_4 := \{a_1, a_3\}, \\[-2pt] %% TYPESETTING
    & \textit{Branch}~\xi_3~\xi_3 := \{a_4\}, \\[-2pt]
    & \textit{Branch}~\xi_3~\xi_4 := \{a_1, a_3, a_4\})\end{aligned}$ \\
\hbox{}\qquad\qquad $\lambda x_1.\> \textit{swap}~x_1~a~b = \undef
    (\!\begin{aligned}[t]%
    & \xi_3 := \xi_3,\> \xi_4 := \xi_3, \\[-2pt]
    & \textit{Branch}~\xi_3~\xi_3 := \textit{Branch}~\xi_3~\xi_3, \\[-2pt]
    & \textit{Branch}~\xi_4~\xi_3 := \textit{Branch}~\xi_3~\xi_3)\end{aligned}$ \\[2\smallskipamount]
The existence of a nonstandard model suggests that the induction hypothesis is not general enough or perhaps
even wrong. See the ``Inductive Properties'' section of the Nitpick manual for details (``\textit{isabelle doc nitpick}'').
\postw

Reading the Nitpick manual is a most excellent idea.
But what's going on? The \textit{non\_std} ``${\kern1pt'a}~\textit{bin\_tree}$''
option told the tool to look for nonstandard models of binary trees, which
means that new ``nonstandard'' trees $\xi_1, \xi_2, \ldots$, are now allowed in
addition to the standard trees generated by the \textit{Leaf} and
\textit{Branch} constructors.%
\footnote{Notice the similarity between allowing nonstandard trees here and
allowing unreachable states in the preceding example (by removing the ``$n \in
\textit{reach\/}$'' assumption). In both cases, we effectively enlarge the
set of objects over which the induction is performed while doing the step
so as to test the induction hypothesis's strength.}
The new trees are so nonstandard that we know nothing about them, except what
the induction hypothesis states and what can be proved about all trees without
relying on induction or case distinction. The key observation is,
%
\begin{quote}
\textsl{If the induction
hypothesis is strong enough, the induction step will hold even for nonstandard
objects, and Nitpick won't find any nonstandard counterexample.}
\end{quote}
%
But here, Nitpick did find some nonstandard trees $t = \xi_3$
and $u = \xi_4$ such that $a \in \textit{labels}~t$, $b \notin
\textit{labels}~t$, $a \notin \textit{labels}~u$, and $b \in \textit{labels}~u$.
Because neither tree contains both $a$ and $b$, the induction hypothesis tells
us nothing about the labels of $\textit{swap}~t~a~b$ and $\textit{swap}~u~a~b$,
and as a result we know nothing about the labels of the tree
$\textit{swap}~(\textit{Branch}~t~u)~a~b$, which by definition equals
$\textit{Branch}$ $(\textit{swap}~t~a~b)$ $(\textit{swap}~u~a~b)$, whose
labels are $\textit{labels}$ $(\textit{swap}~t~a~b) \mathrel{\cup}
\textit{labels}$ $(\textit{swap}~u~a~b)$.

The solution is to ensure that we always know what the labels of the subtrees
are in the inductive step, by covering the cases where $a$ and/or~$b$ is not in
$t$ in the statement of the lemma:

\prew
\textbf{lemma} ``$\textit{labels}~(\textit{swap}~t~a~b) = {}$ \\
\phantom{\textbf{lemma} ``}$(\textrm{if}~a \in \textit{labels}~t~\textrm{then}$ \nopagebreak \\
\phantom{\textbf{lemma} ``(\quad}$\textrm{if}~b \in \textit{labels}~t~\textrm{then}~\textit{labels}~t~\textrm{else}~(\textit{labels}~t - \{a\}) \mathrel{\cup} \{b\}$ \\
\phantom{\textbf{lemma} ``(}$\textrm{else}$ \\
\phantom{\textbf{lemma} ``(\quad}$\textrm{if}~b \in \textit{labels}~t~\textrm{then}~(\textit{labels}~t - \{b\}) \mathrel{\cup} \{a\}~\textrm{else}~\textit{labels}~t)$''
\postw