\documentclass[a4paper,12pt]{article}

\usepackage[T1]{fontenc}

\usepackage{amsmath}

\usepackage{amssymb}

\usepackage[english,french]{babel}

\usepackage{color}

\usepackage{graphicx}

%\usepackage{mathpazo}

\usepackage{multicol}

\usepackage{stmaryrd}

%\usepackage[scaled=.85]{beramono}

\usepackage{../iman,../pdfsetup}

%\oddsidemargin=4.6mm

%\evensidemargin=4.6mm

%\textwidth=150mm

%\topmargin=4.6mm

%\headheight=0mm

%\headsep=0mm

%\textheight=234mm

\def\Colon{\mathord{:\mkern-1.5mu:}}

%\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\undef{\textit{undefined}}

\def\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 2010}

\author{\hbox{} \\

Jasmin Christian Blanchette \\

{\normalsize Fakult\"at f\"ur Informatik, Technische Universit\"at M\"unchen} \\

\hbox{}}

\maketitle

\tableofcontents

\setlength{\parskip}{.7em plus .2em minus .1em}

\setlength{\parindent}{0pt}

\setlength{\abovedisplayskip}{\parskip}

\setlength{\abovedisplayshortskip}{.9\parskip}

\setlength{\belowdisplayskip}{\parskip}

\setlength{\belowdisplayshortskip}{.9\parskip}

% General-purpose enum environment with correct spacing

\newenvironment{enum}%

    {\begin{list}{}{%

        \setlength{\topsep}{.1\parskip}%

        \setlength{\partopsep}{.1\parskip}%

        \setlength{\itemsep}{\parskip}%

        \advance\itemsep by-\parsep}}

    {\end{list}}

\def\pre{\begingroup\vskip0pt plus1ex\advance\leftskip by\leftmargin

\advance\rightskip by\leftmargin}

\def\post{\vskip0pt plus1ex\endgroup}

\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): \\

\ 1. $P\,\Longrightarrow\, Q$ \\

\ 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 \undef{} symbol's presence is explained as

follows: In higher-order logic, any function can be built from the undefined

function using repeated applications of the function update operator $f(x :=

y)$, just like any list can be built from the empty list using $x \mathbin{\#}

xs$.}

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 considers prefixes $\{0,\, 1,\,

\ldots,\, K - 1\}$ of \textit{nat} (where $K = \textit{card}~\textit{nat}$) and

maps all other numbers to the undefined value ($\unk$). The type \textit{int} is

handled in a similar way: If $K = \textit{card}~\textit{int}$, the subset of

\textit{int} known to Nitpick is $\{-\lceil K/2 \rceil + 1,\, \ldots,\, +\lfloor

K/2 \rfloor\}$. 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

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 the option \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 often 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

For any cardinality $k$, \textit{Suc} is the partial function $\{0 \mapsto 1,\,

1 \mapsto 2,\, \ldots,\, k - 1 \mapsto \unk\}$, which evaluates to $\unk$ when

it is 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 $\lambda x_1.\; x_1 \mathbin{@} [y, y] = \undef([] := [a_3, a_3],\> [a_3, a_3] := \unk,\> [a_3] := \unk)$ \\

\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{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] \\[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]