doc-src/ZF/FOL.tex
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%!TEX root = logics-ZF.tex
\chapter{First-Order Logic}
\index{first-order logic|(}

Isabelle implements Gentzen's natural deduction systems {\sc nj} and {\sc
nk}.  Intuitionistic first-order logic is defined first, as theory
\thydx{IFOL}.  Classical logic, theory \thydx{FOL}, is
obtained by adding the double negation rule.  Basic proof procedures are
provided.  The intuitionistic prover works with derived rules to simplify
implications in the assumptions.  Classical~\texttt{FOL} employs Isabelle's
classical reasoner, which simulates a sequent calculus.

\section{Syntax and rules of inference}
The logic is many-sorted, using Isabelle's type classes.  The class of
first-order terms is called \cldx{term} and is a subclass of \isa{logic}.
No types of individuals are provided, but extensions can define types such
as \isa{nat::term} and type constructors such as \isa{list::(term)term}
(see the examples directory, \texttt{FOL/ex}).  Below, the type variable
$\alpha$ ranges over class \isa{term}; the equality symbol and quantifiers
are polymorphic (many-sorted).  The type of formulae is~\tydx{o}, which
belongs to class~\cldx{logic}.  Figure~\ref{fol-syntax} gives the syntax.
Note that $a$\verb|~=|$b$ is translated to $\neg(a=b)$.

Figure~\ref{fol-rules} shows the inference rules with their names.
Negation is defined in the usual way for intuitionistic logic; $\neg P$
abbreviates $P\imp\bot$.  The biconditional~($\bimp$) is defined through
$\conj$ and~$\imp$; introduction and elimination rules are derived for it.

The unique existence quantifier, $\exists!x.P(x)$, is defined in terms
of~$\exists$ and~$\forall$.  An Isabelle binder, it admits nested
quantifications.  For instance, $\exists!x\;y.P(x,y)$ abbreviates
$\exists!x. \exists!y.P(x,y)$; note that this does not mean that there
exists a unique pair $(x,y)$ satisfying~$P(x,y)$.

Some intuitionistic derived rules are shown in
Fig.\ts\ref{fol-int-derived}, again with their names.  These include
rules for the defined symbols $\neg$, $\bimp$ and $\exists!$.  Natural
deduction typically involves a combination of forward and backward
reasoning, particularly with the destruction rules $(\conj E)$,
$({\imp}E)$, and~$(\forall E)$.  Isabelle's backward style handles these
rules badly, so sequent-style rules are derived to eliminate conjunctions,
implications, and universal quantifiers.  Used with elim-resolution,
\tdx{allE} eliminates a universal quantifier while \tdx{all_dupE}
re-inserts the quantified formula for later use.  The rules
\isa{conj\_impE}, etc., support the intuitionistic proof procedure
(see~\S\ref{fol-int-prover}).

See the on-line theory library for complete listings of the rules and
derived rules.

\begin{figure}
\begin{center}
\begin{tabular}{rrr}
\it name      &\it meta-type  & \it description \\
\cdx{Trueprop}& $o\To prop$           & coercion to $prop$\\
\cdx{Not}     & $o\To o$              & negation ($\neg$) \\
\cdx{True}    & $o$                   & tautology ($\top$) \\
\cdx{False}   & $o$                   & absurdity ($\bot$)
\end{tabular}
\end{center}
\subcaption{Constants}

\begin{center}
\begin{tabular}{llrrr}
\it symbol &\it name     &\it meta-type & \it priority & \it description \\
\sdx{ALL}  & \cdx{All}  & $(\alpha\To o)\To o$ & 10 &
universal quantifier ($\forall$) \\
\sdx{EX}   & \cdx{Ex}   & $(\alpha\To o)\To o$ & 10 &
existential quantifier ($\exists$) \\
\isa{EX!}  & \cdx{Ex1}  & $(\alpha\To o)\To o$ & 10 &
unique existence ($\exists!$)
\end{tabular}
\index{*"E"X"! symbol}
\end{center}
\subcaption{Binders}

\begin{center}
\index{*"= symbol}
\index{&@{\tt\&} symbol}
\index{*"| symbol}
\index{*"-"-"> symbol}
\index{*"<"-"> symbol}
\begin{tabular}{rrrr}
\it symbol    & \it meta-type         & \it priority & \it description \\
\tt =         & $[\alpha,\alpha]\To o$ & Left 50 & equality ($=$) \\
\tt \&        & $[o,o]\To o$          & Right 35 & conjunction ($\conj$) \\
\tt |         & $[o,o]\To o$          & Right 30 & disjunction ($\disj$) \\
\tt -->       & $[o,o]\To o$          & Right 25 & implication ($\imp$) \\
\tt <->       & $[o,o]\To o$          & Right 25 & biconditional ($\bimp$)
\end{tabular}
\end{center}
\subcaption{Infixes}

\dquotes
$\begin{array}{rcl} formula & = & \hbox{expression of type~o} \\ & | & term " = " term \quad| \quad term " \ttilde= " term \\ & | & "\ttilde\ " formula \\ & | & formula " \& " formula \\ & | & formula " | " formula \\ & | & formula " --> " formula \\ & | & formula " <-> " formula \\ & | & "ALL~" id~id^* " . " formula \\ & | & "EX~~" id~id^* " . " formula \\ & | & "EX!~" id~id^* " . " formula \end{array}$
\subcaption{Grammar}
\caption{Syntax of \texttt{FOL}} \label{fol-syntax}
\end{figure}

\begin{figure}
\begin{ttbox}
\tdx{refl}        a=a
\tdx{subst}       [| a=b;  P(a) |] ==> P(b)
\subcaption{Equality rules}

\tdx{conjI}       [| P;  Q |] ==> P&Q
\tdx{conjunct1}   P&Q ==> P
\tdx{conjunct2}   P&Q ==> Q

\tdx{disjI1}      P ==> P|Q
\tdx{disjI2}      Q ==> P|Q
\tdx{disjE}       [| P|Q;  P ==> R;  Q ==> R |] ==> R

\tdx{impI}        (P ==> Q) ==> P-->Q
\tdx{mp}          [| P-->Q;  P |] ==> Q

\tdx{FalseE}      False ==> P
\subcaption{Propositional rules}

\tdx{allI}        (!!x. P(x))  ==> (ALL x.P(x))
\tdx{spec}        (ALL x.P(x)) ==> P(x)

\tdx{exI}         P(x) ==> (EX x.P(x))
\tdx{exE}         [| EX x.P(x);  !!x. P(x) ==> R |] ==> R
\subcaption{Quantifier rules}

\tdx{True_def}    True        == False-->False
\tdx{not_def}     ~P          == P-->False
\tdx{iff_def}     P<->Q       == (P-->Q) & (Q-->P)
\tdx{ex1_def}     EX! x. P(x) == EX x. P(x) & (ALL y. P(y) --> y=x)
\subcaption{Definitions}
\end{ttbox}

\caption{Rules of intuitionistic logic} \label{fol-rules}
\end{figure}

\begin{figure}
\begin{ttbox}
\tdx{sym}       a=b ==> b=a
\tdx{trans}     [| a=b;  b=c |] ==> a=c
\tdx{ssubst}    [| b=a;  P(a) |] ==> P(b)
\subcaption{Derived equality rules}

\tdx{TrueI}     True

\tdx{notI}      (P ==> False) ==> ~P
\tdx{notE}      [| ~P;  P |] ==> R

\tdx{iffI}      [| P ==> Q;  Q ==> P |] ==> P<->Q
\tdx{iffE}      [| P <-> Q;  [| P-->Q; Q-->P |] ==> R |] ==> R
\tdx{iffD1}     [| P <-> Q;  P |] ==> Q
\tdx{iffD2}     [| P <-> Q;  Q |] ==> P

\tdx{ex1I}      [| P(a);  !!x. P(x) ==> x=a |]  ==>  EX! x. P(x)
\tdx{ex1E}      [| EX! x.P(x);  !!x.[| P(x);  ALL y. P(y) --> y=x |] ==> R
|] ==> R
\subcaption{Derived rules for $$\top$$, $$\neg$$, $$\bimp$$ and $$\exists!$$}

\tdx{conjE}     [| P&Q;  [| P; Q |] ==> R |] ==> R
\tdx{impE}      [| P-->Q;  P;  Q ==> R |] ==> R
\tdx{allE}      [| ALL x.P(x);  P(x) ==> R |] ==> R
\tdx{all_dupE}  [| ALL x.P(x);  [| P(x); ALL x.P(x) |] ==> R |] ==> R
\subcaption{Sequent-style elimination rules}

\tdx{conj_impE} [| (P&Q)-->S;  P-->(Q-->S) ==> R |] ==> R
\tdx{disj_impE} [| (P|Q)-->S;  [| P-->S; Q-->S |] ==> R |] ==> R
\tdx{imp_impE}  [| (P-->Q)-->S;  [| P; Q-->S |] ==> Q;  S ==> R |] ==> R
\tdx{not_impE}  [| ~P --> S;  P ==> False;  S ==> R |] ==> R
\tdx{iff_impE}  [| (P<->Q)-->S; [| P; Q-->S |] ==> Q; [| Q; P-->S |] ==> P;
S ==> R |] ==> R
\tdx{all_impE}  [| (ALL x.P(x))-->S;  !!x.P(x);  S ==> R |] ==> R
\tdx{ex_impE}   [| (EX x.P(x))-->S;  P(a)-->S ==> R |] ==> R
\end{ttbox}
\subcaption{Intuitionistic simplification of implication}
\caption{Derived rules for intuitionistic logic} \label{fol-int-derived}
\end{figure}

\section{Generic packages}
FOL instantiates most of Isabelle's generic packages.
\begin{itemize}
\item
It instantiates the simplifier, which is invoked through the method
\isa{simp}.  Both equality ($=$) and the biconditional
($\bimp$) may be used for rewriting.

\item
It instantiates the classical reasoner, which is invoked mainly through the
methods \isa{blast} and \isa{auto}.  See~\S\ref{fol-cla-prover}
for details.
%
%\item FOL provides the tactic \ttindex{hyp_subst_tac}, which substitutes for
%  an equality throughout a subgoal and its hypotheses.  This tactic uses FOL's
%  general substitution rule.
\end{itemize}

\begin{warn}\index{simplification!of conjunctions}%
Simplifying $a=b\conj P(a)$ to $a=b\conj P(b)$ is often advantageous.  The
left part of a conjunction helps in simplifying the right part.  This effect
is not available by default: it can be slow.  It can be obtained by
including the theorem \isa{conj_cong}\index{*conj_cong (rule)}%
as a congruence rule in
simplification, \isa{simp cong:\ conj\_cong}.
\end{warn}

\section{Intuitionistic proof procedures} \label{fol-int-prover}
Implication elimination (the rules~\isa{mp} and~\isa{impE}) pose
difficulties for automated proof.  In intuitionistic logic, the assumption
$P\imp Q$ cannot be treated like $\neg P\disj Q$.  Given $P\imp Q$, we may
use~$Q$ provided we can prove~$P$; the proof of~$P$ may require repeated
use of $P\imp Q$.  If the proof of~$P$ fails then the whole branch of the
proof must be abandoned.  Thus intuitionistic propositional logic requires
backtracking.

For an elementary example, consider the intuitionistic proof of $Q$ from
$P\imp Q$ and $(P\imp Q)\imp P$.  The implication $P\imp Q$ is needed
twice:
$\infer[({\imp}E)]{Q}{P\imp Q & \infer[({\imp}E)]{P}{(P\imp Q)\imp P & P\imp Q}}$
The theorem prover for intuitionistic logic does not use~\isa{impE}.\@
Instead, it simplifies implications using derived rules
(Fig.\ts\ref{fol-int-derived}).  It reduces the antecedents of implications
to atoms and then uses Modus Ponens: from $P\imp Q$ and~$P$ deduce~$Q$.
The rules \tdx{conj_impE} and \tdx{disj_impE} are
straightforward: $(P\conj Q)\imp S$ is equivalent to $P\imp (Q\imp S)$, and
$(P\disj Q)\imp S$ is equivalent to the conjunction of $P\imp S$ and $Q\imp S$.  The other \ldots\isa{\_impE} rules are unsafe; the method requires
backtracking.  All the rules are derived in the same simple manner.

Dyckhoff has independently discovered similar rules, and (more importantly)
has demonstrated their completeness for propositional
logic~\cite{dyckhoff}.  However, the tactics given below are not complete
for first-order logic because they discard universally quantified
assumptions after a single use. These are \ML{} functions, and are listed
below with their \ML{} type:
\begin{ttbox}
mp_tac              : int -> tactic
eq_mp_tac           : int -> tactic
IntPr.safe_step_tac : int -> tactic
IntPr.safe_tac      :        tactic
IntPr.inst_step_tac : int -> tactic
IntPr.step_tac      : int -> tactic
IntPr.fast_tac      : int -> tactic
IntPr.best_tac      : int -> tactic
\end{ttbox}
Most of these belong to the structure \ML{} structure \texttt{IntPr} and resemble the
tactics of Isabelle's classical reasoner.  There are no corresponding
Isar methods, but you can use the
\isa{tactic} method to call \ML{} tactics in an Isar script:
\begin{isabelle}
\isacommand{apply}\ (tactic\ {* IntPr.fast\_tac 1*})
\end{isabelle}
Here is a description of each tactic:

\begin{ttdescription}
\item[\ttindexbold{mp_tac} {\it i}]
attempts to use \tdx{notE} or \tdx{impE} within the assumptions in
subgoal $i$.  For each assumption of the form $\neg P$ or $P\imp Q$, it
searches for another assumption unifiable with~$P$.  By
contradiction with $\neg P$ it can solve the subgoal completely; by Modus
Ponens it can replace the assumption $P\imp Q$ by $Q$.  The tactic can
produce multiple outcomes, enumerating all suitable pairs of assumptions.

\item[\ttindexbold{eq_mp_tac} {\it i}]
is like \texttt{mp_tac} {\it i}, but may not instantiate unknowns --- thus, it
is safe.

\item[\ttindexbold{IntPr.safe_step_tac} $i$] performs a safe step on
subgoal~$i$.  This may include proof by assumption or Modus Ponens (taking
care not to instantiate unknowns), or \texttt{hyp_subst_tac}.

\item[\ttindexbold{IntPr.safe_tac}] repeatedly performs safe steps on all
subgoals.  It is deterministic, with at most one outcome.

\item[\ttindexbold{IntPr.inst_step_tac} $i$] is like \texttt{safe_step_tac},
but allows unknowns to be instantiated.

\item[\ttindexbold{IntPr.step_tac} $i$] tries \texttt{safe_tac} or {\tt
inst_step_tac}, or applies an unsafe rule.  This is the basic step of
the intuitionistic proof procedure.

\item[\ttindexbold{IntPr.fast_tac} $i$] applies \texttt{step_tac}, using
depth-first search, to solve subgoal~$i$.

\item[\ttindexbold{IntPr.best_tac} $i$] applies \texttt{step_tac}, using
best-first search (guided by the size of the proof state) to solve subgoal~$i$.
\end{ttdescription}
Here are some of the theorems that \texttt{IntPr.fast_tac} proves
automatically.  The latter three date from {\it Principia Mathematica}
(*11.53, *11.55, *11.61)~\cite{principia}.
\begin{ttbox}
~~P & ~~(P --> Q) --> ~~Q
(ALL x y. P(x) --> Q(y)) <-> ((EX x. P(x)) --> (ALL y. Q(y)))
(EX x y. P(x) & Q(x,y)) <-> (EX x. P(x) & (EX y. Q(x,y)))
(EX y. ALL x. P(x) --> Q(x,y)) --> (ALL x. P(x) --> (EX y. Q(x,y)))
\end{ttbox}

\begin{figure}
\begin{ttbox}
\tdx{excluded_middle}    ~P | P

\tdx{disjCI}    (~Q ==> P) ==> P|Q
\tdx{exCI}      (ALL x. ~P(x) ==> P(a)) ==> EX x.P(x)
\tdx{impCE}     [| P-->Q; ~P ==> R; Q ==> R |] ==> R
\tdx{iffCE}     [| P<->Q;  [| P; Q |] ==> R;  [| ~P; ~Q |] ==> R |] ==> R
\tdx{notnotD}   ~~P ==> P
\tdx{swap}      ~P ==> (~Q ==> P) ==> Q
\end{ttbox}
\caption{Derived rules for classical logic} \label{fol-cla-derived}
\end{figure}

\section{Classical proof procedures} \label{fol-cla-prover}
The classical theory, \thydx{FOL}, consists of intuitionistic logic plus
the rule
$$\vcenter{\infer{P}{\infer*{P}{[\neg P]}}} \eqno(classical)$$
\noindent
Natural deduction in classical logic is not really all that natural.  FOL
derives classical introduction rules for $\disj$ and~$\exists$, as well as
classical elimination rules for~$\imp$ and~$\bimp$, and the swap rule (see
Fig.\ts\ref{fol-cla-derived}).

The classical reasoner is installed.  The most useful methods are
\isa{blast} (pure classical reasoning) and \isa{auto} (classical reasoning
combined with simplification), but the full range of
methods is available, including \isa{clarify},
\isa{fast}, \isa{best} and \isa{force}.
See the
\iflabelundefined{chap:classical}{the {\em Reference Manual\/}}%
{Chap.\ts\ref{chap:classical}}
and the \emph{Tutorial}~\cite{isa-tutorial}
for more discussion of classical proof methods.

\section{An intuitionistic example}
Here is a session similar to one in the book {\em Logic and Computation}
\cite[pages~222--3]{paulson87}. It illustrates the treatment of
quantifier reasoning, which is different in Isabelle than it is in
{\sc lcf}-based theorem provers such as {\sc hol}.

The theory header specifies that we are working in intuitionistic
logic by designating \isa{IFOL} rather than \isa{FOL} as the parent
theory:
\begin{isabelle}
\isacommand{theory}\ IFOL\_examples\ \isacommand{imports}\ IFOL\isanewline
\isacommand{begin}
\end{isabelle}
The proof begins by entering the goal, then applying the rule $({\imp}I)$.
\begin{isabelle}
\isacommand{lemma}\ "(EX\ y.\ ALL\ x.\ Q(x,y))\ -->\ \ (ALL\ x.\ EX\ y.\ Q(x,y))"\isanewline
\ 1.\ (\isasymexists y.\ \isasymforall x.\ Q(x,\ y))\
\isasymlongrightarrow \ (\isasymforall x.\ \isasymexists y.\ Q(x,\ y))
\isanewline
\isacommand{apply}\ (rule\ impI)\isanewline
\ 1.\ \isasymexists y.\ \isasymforall x.\ Q(x,\ y)\
\isasymLongrightarrow \ \isasymforall x.\ \isasymexists y.\ Q(x,\ y)
\end{isabelle}
Isabelle's output is shown as it would appear using Proof General.
In this example, we shall never have more than one subgoal.  Applying
$({\imp}I)$ replaces~\isa{\isasymlongrightarrow}
by~\isa{\isasymLongrightarrow}, so that
$$\ex{y}\all{x}Q(x,y)$$ becomes an assumption.  We have the choice of
$({\exists}E)$ and $({\forall}I)$; let us try the latter.
\begin{isabelle}
\isacommand{apply}\ (rule\ allI)\isanewline
\ 1.\ \isasymAnd x.\ \isasymexists y.\ \isasymforall x.\ Q(x,\ y)\
\isasymLongrightarrow \ \isasymexists y.\ Q(x,\ y)\hfill$$(*)$$
\end{isabelle}
Applying $({\forall}I)$ replaces the \isa{\isasymforall
x} (in ASCII, \isa{ALL~x}) by \isa{\isasymAnd x}
(or \isa{!!x}), replacing FOL's universal quantifier by Isabelle's
meta universal quantifier.  The bound variable is a {\bf parameter} of
the subgoal.  We now must choose between $({\exists}I)$ and
$({\exists}E)$.  What happens if the wrong rule is chosen?
\begin{isabelle}
\isacommand{apply}\ (rule\ exI)\isanewline
\ 1.\ \isasymAnd x.\ \isasymexists y.\ \isasymforall x.\ Q(x,\ y)\
\isasymLongrightarrow \ Q(x,\ ?y2(x))
\end{isabelle}
The new subgoal~1 contains the function variable \isa{?y2}.  Instantiating
\isa{?y2} can replace~\isa{?y2(x)} by a term containing~\isa{x}, even
though~\isa{x} is a bound variable.  Now we analyse the assumption
$$\exists y.\forall x. Q(x,y)$$ using elimination rules:
\begin{isabelle}
\isacommand{apply}\ (erule\ exE)\isanewline
\ 1.\ \isasymAnd x\ y.\ \isasymforall x.\ Q(x,\ y)\ \isasymLongrightarrow \ Q(x,\ ?y2(x))
\end{isabelle}
Applying $(\exists E)$ has produced the parameter \isa{y} and stripped the
existential quantifier from the assumption.  But the subgoal is unprovable:
there is no way to unify \isa{?y2(x)} with the bound variable~\isa{y}.
$(*)$ above.  This time we apply $({\exists}E)$:
\begin{isabelle}
\isacommand{apply}\ (erule\ exE)\isanewline
\ 1.\ \isasymAnd x\ y.\ \isasymforall x.\ Q(x,\ y)\
\isasymLongrightarrow \ \isasymexists y.\ Q(x,\ y)
\end{isabelle}
We now have two parameters and no scheme variables.  Applying
$({\exists}I)$ and $({\forall}E)$ produces two scheme variables, which are
applied to those parameters.  Parameters should be produced early, as this
example demonstrates.
\begin{isabelle}
\isacommand{apply}\ (rule\ exI)\isanewline
\ 1.\ \isasymAnd x\ y.\ \isasymforall x.\ Q(x,\ y)\
\isasymLongrightarrow \ Q(x,\ ?y3(x,\ y))
\isanewline
\isacommand{apply}\ (erule\ allE)\isanewline
\ 1.\ \isasymAnd x\ y.\ Q(?x4(x,\ y),\ y)\ \isasymLongrightarrow \
Q(x,\ ?y3(x,\ y))
\end{isabelle}
The subgoal has variables \isa{?y3} and \isa{?x4} applied to both
parameters.  The obvious projection functions unify \isa{?x4(x,y)} with~\isa{
x} and \isa{?y3(x,y)} with~\isa{y}.
\begin{isabelle}
\isacommand{apply}\ assumption\isanewline
No\ subgoals!\isanewline
\isacommand{done}
\end{isabelle}
The theorem was proved in six method invocations, not counting the
abandoned ones.  But proof checking is tedious, and the \ML{} tactic
\ttindex{IntPr.fast_tac} can prove the theorem in one step.
\begin{isabelle}
\isacommand{lemma}\ "(EX\ y.\ ALL\ x.\ Q(x,y))\ -->\ \ (ALL\ x.\ EX\ y.\ Q(x,y))"\isanewline
\ 1.\ (\isasymexists y.\ \isasymforall x.\ Q(x,\ y))\
\isasymlongrightarrow \ (\isasymforall x.\ \isasymexists y.\ Q(x,\ y))
\isanewline
\isacommand{by} (tactic \ttlbrace*IntPr.fast_tac 1*\ttrbrace)\isanewline
No\ subgoals!
\end{isabelle}

\section{An example of intuitionistic negation}
The following example demonstrates the specialized forms of implication
elimination.  Even propositional formulae can be difficult to prove from
the basic rules; the specialized rules help considerably.

Propositional examples are easy to invent.  As Dummett notes~\cite[page
28]{dummett}, $\neg P$ is classically provable if and only if it is
intuitionistically provable;  therefore, $P$ is classically provable if and
only if $\neg\neg P$ is intuitionistically provable.%
\footnote{This remark holds only for propositional logic, not if $P$ is
allowed to contain quantifiers.}
%
Proving $\neg\neg P$ intuitionistically is
much harder than proving~$P$ classically.

Our example is the double negation of the classical tautology $(P\imp Q)\disj (Q\imp P)$.  The first step is apply the
\isa{unfold} method, which expands
negations to implications using the definition $\neg P\equiv P\imp\bot$
and allows use of the special implication rules.
\begin{isabelle}
\isacommand{lemma}\ "\isachartilde \ \isachartilde \ ((P-->Q)\ |\ (Q-->P))"\isanewline
\ 1.\ \isasymnot \ \isasymnot \ ((P\ \isasymlongrightarrow \ Q)\ \isasymor \ (Q\ \isasymlongrightarrow \ P))
\isanewline
\isacommand{apply}\ (unfold\ not\_def)\isanewline
\ 1.\ ((P\ \isasymlongrightarrow \ Q)\ \isasymor \ (Q\ \isasymlongrightarrow \ P)\ \isasymlongrightarrow \ False)\ \isasymlongrightarrow \ False%
\end{isabelle}
The next step is a trivial use of $(\imp I)$.
\begin{isabelle}
\isacommand{apply}\ (rule\ impI)\isanewline
\ 1.\ (P\ \isasymlongrightarrow \ Q)\ \isasymor \ (Q\ \isasymlongrightarrow \ P)\ \isasymlongrightarrow \ False\ \isasymLongrightarrow \ False%
\end{isabelle}
By $(\imp E)$ it would suffice to prove $(P\imp Q)\disj (Q\imp P)$, but
that formula is not a theorem of intuitionistic logic.  Instead, we
apply the specialized implication rule \tdx{disj_impE}.  It splits the
assumption into two assumptions, one for each disjunct.
\begin{isabelle}
\isacommand{apply}\ (erule\ disj\_impE)\isanewline
\ 1.\ \isasymlbrakk (P\ \isasymlongrightarrow \ Q)\ \isasymlongrightarrow \ False;\ (Q\ \isasymlongrightarrow \ P)\ \isasymlongrightarrow \ False\isasymrbrakk \ \isasymLongrightarrow \
False
\end{isabelle}
We cannot hope to prove $P\imp Q$ or $Q\imp P$ separately, but
their negations are inconsistent.  Applying \tdx{imp_impE} breaks down
the assumption $\neg(P\imp Q)$, asking to show~$Q$ while providing new
assumptions~$P$ and~$\neg Q$.
\begin{isabelle}
\isacommand{apply}\ (erule\ imp\_impE)\isanewline
\ 1.\ \isasymlbrakk (Q\ \isasymlongrightarrow \ P)\ \isasymlongrightarrow \ False;\ P;\ Q\ \isasymlongrightarrow \ False\isasymrbrakk \ \isasymLongrightarrow \ Q\isanewline
\ 2.\ \isasymlbrakk (Q\ \isasymlongrightarrow \ P)\ \isasymlongrightarrow \ False;\ False\isasymrbrakk \ \isasymLongrightarrow \
False
\end{isabelle}
Subgoal~2 holds trivially; let us ignore it and continue working on
subgoal~1.  Thanks to the assumption~$P$, we could prove $Q\imp P$;
applying \tdx{imp_impE} is simpler.
\begin{isabelle}
\ \isacommand{apply}\ (erule\ imp\_impE)\isanewline
\ 1.\ \isasymlbrakk P;\ Q\ \isasymlongrightarrow \ False;\ Q;\ P\ \isasymlongrightarrow \ False\isasymrbrakk \ \isasymLongrightarrow \ P\isanewline
\ 2.\ \isasymlbrakk P;\ Q\ \isasymlongrightarrow \ False;\ False\isasymrbrakk \ \isasymLongrightarrow \ Q\isanewline
\ 3.\ \isasymlbrakk (Q\ \isasymlongrightarrow \ P)\ \isasymlongrightarrow \ False;\ False\isasymrbrakk \ \isasymLongrightarrow \ False%
\end{isabelle}
The three subgoals are all trivial.
\begin{isabelle}
\isacommand{apply}\ assumption\ \isanewline
\ 1.\ \isasymlbrakk P;\ Q\ \isasymlongrightarrow \ False;\
False\isasymrbrakk \ \isasymLongrightarrow \ Q\isanewline
\ 2.\ \isasymlbrakk (Q\ \isasymlongrightarrow \ P)\
\isasymlongrightarrow \ False;\ False\isasymrbrakk \
\isasymLongrightarrow \ False%
\isanewline
\isacommand{apply}\ (erule\ FalseE)+\isanewline
No\ subgoals!\isanewline
\isacommand{done}
\end{isabelle}
This proof is also trivial for the \ML{} tactic \isa{IntPr.fast_tac}.

\section{A classical example} \label{fol-cla-example}
To illustrate classical logic, we shall prove the theorem
$\ex{y}\all{x}P(y)\imp P(x)$.  Informally, the theorem can be proved as
follows.  Choose~$y$ such that~$\neg P(y)$, if such exists; otherwise
$\all{x}P(x)$ is true.  Either way the theorem holds.  First, we must
work in a theory based on classical logic, the theory \isa{FOL}:
\begin{isabelle}
\isacommand{theory}\ FOL\_examples\ \isacommand{imports}\ FOL\isanewline
\isacommand{begin}
\end{isabelle}

The formal proof does not conform in any obvious way to the sketch given
above.  Its key step is its first rule, \tdx{exCI}, a classical
version of~$(\exists I)$ that allows multiple instantiation of the
quantifier.
\begin{isabelle}
\isacommand{lemma}\ "EX\ y.\ ALL\ x.\ P(y)-->P(x)"\isanewline
\ 1.\ \isasymexists y.\ \isasymforall x.\ P(y)\ \isasymlongrightarrow \ P(x)
\isanewline
\isacommand{apply}\ (rule\ exCI)\isanewline
\ 1.\ \isasymforall y.\ \isasymnot \ (\isasymforall x.\ P(y)\ \isasymlongrightarrow \ P(x))\ \isasymLongrightarrow \ \isasymforall x.\ P(?a)\ \isasymlongrightarrow \ P(x)
\end{isabelle}
We can either exhibit a term \isa{?a} to satisfy the conclusion of
subgoal~1, or produce a contradiction from the assumption.  The next
steps are routine.
\begin{isabelle}
\isacommand{apply}\ (rule\ allI)\isanewline
\ 1.\ \isasymAnd x.\ \isasymforall y.\ \isasymnot \ (\isasymforall x.\ P(y)\ \isasymlongrightarrow \ P(x))\ \isasymLongrightarrow \ P(?a)\ \isasymlongrightarrow \ P(x)
\isanewline
\isacommand{apply}\ (rule\ impI)\isanewline
\ 1.\ \isasymAnd x.\ \isasymlbrakk \isasymforall y.\ \isasymnot \ (\isasymforall x.\ P(y)\ \isasymlongrightarrow \ P(x));\ P(?a)\isasymrbrakk \ \isasymLongrightarrow \ P(x)
\end{isabelle}
By the duality between $\exists$ and~$\forall$, applying~$(\forall E)$
is equivalent to applying~$(\exists I)$ again.
\begin{isabelle}
\isacommand{apply}\ (erule\ allE)\isanewline
\ 1.\ \isasymAnd x.\ \isasymlbrakk P(?a);\ \isasymnot \ (\isasymforall xa.\ P(?y3(x))\ \isasymlongrightarrow \ P(xa))\isasymrbrakk \ \isasymLongrightarrow \ P(x)
\end{isabelle}
In classical logic, a negated assumption is equivalent to a conclusion.  To
get this effect, we create a swapped version of $(\forall I)$ and apply it
using \ttindex{erule}.
\begin{isabelle}
\isacommand{apply}\ (erule\ allI\ [THEN\ [2]\ swap])\isanewline
\ 1.\ \isasymAnd x\ xa.\ \isasymlbrakk P(?a);\ \isasymnot \ P(x)\isasymrbrakk \ \isasymLongrightarrow \ P(?y3(x))\ \isasymlongrightarrow \ P(xa)
\end{isabelle}
The operand of \isa{erule} above denotes the following theorem:
\begin{isabelle}
\ \ \ \ \isasymlbrakk \isasymnot \ (\isasymforall x.\ ?P1(x));\
\isasymAnd x.\ \isasymnot \ ?P\ \isasymLongrightarrow \
?P1(x)\isasymrbrakk \
\isasymLongrightarrow \ ?P%
\end{isabelle}
The previous conclusion, \isa{P(x)}, has become a negated assumption.
\begin{isabelle}
\isacommand{apply}\ (rule\ impI)\isanewline
\ 1.\ \isasymAnd x\ xa.\ \isasymlbrakk P(?a);\ \isasymnot \ P(x);\ P(?y3(x))\isasymrbrakk \ \isasymLongrightarrow \ P(xa)
\end{isabelle}
The subgoal has three assumptions.  We produce a contradiction between the
and~\isa{P(?y3(x))}.   The proof never instantiates the
unknown~\isa{?a}.
\begin{isabelle}
\isacommand{apply}\ (erule\ notE)\isanewline
\ 1.\ \isasymAnd x\ xa.\ \isasymlbrakk P(?a);\ P(?y3(x))\isasymrbrakk \ \isasymLongrightarrow \ P(x)
\isanewline
\isacommand{apply}\ assumption\isanewline
No\ subgoals!\isanewline
\isacommand{done}
\end{isabelle}
The civilised way to prove this theorem is using the
\methdx{blast} method, which automatically uses the classical form
of the rule~$(\exists I)$:
\begin{isabelle}
\isacommand{lemma}\ "EX\ y.\ ALL\ x.\ P(y)-->P(x)"\isanewline
\ 1.\ \isasymexists y.\ \isasymforall x.\ P(y)\ \isasymlongrightarrow \ P(x)
\isanewline
\isacommand{by}\ blast\isanewline
No\ subgoals!
\end{isabelle}
If this theorem seems counterintuitive, then perhaps you are an
intuitionist.  In constructive logic, proving $\ex{y}\all{x}P(y)\imp P(x)$
requires exhibiting a particular term~$t$ such that $\all{x}P(t)\imp P(x)$,
which we cannot do without further knowledge about~$P$.

\section{Derived rules and the classical tactics}
Classical first-order logic can be extended with the propositional
connective $if(P,Q,R)$, where
$$if(P,Q,R) \equiv P\conj Q \disj \neg P \conj R. \eqno(if)$$
Theorems about $if$ can be proved by treating this as an abbreviation,
replacing $if(P,Q,R)$ by $P\conj Q \disj \neg P \conj R$ in subgoals.  But
this duplicates~$P$, causing an exponential blowup and an unreadable
formula.  Introducing further abbreviations makes the problem worse.

Natural deduction demands rules that introduce and eliminate $if(P,Q,R)$
directly, without reference to its definition.  The simple identity
$if(P,Q,R) \,\bimp\, (P\imp Q)\conj (\neg P\imp R)$
suggests that the
$if$-introduction rule should be
$\infer[({if}\,I)]{if(P,Q,R)}{\infer*{Q}{[P]} & \infer*{R}{[\neg P]}}$
The $if$-elimination rule reflects the definition of $if(P,Q,R)$ and the
elimination rules for~$\disj$ and~$\conj$.
$\infer[({if}\,E)]{S}{if(P,Q,R) & \infer*{S}{[P,Q]} & \infer*{S}{[\neg P,R]}}$
Having made these plans, we get down to work with Isabelle.  The theory of
classical logic, \texttt{FOL}, is extended with the constant
$if::[o,o,o]\To o$.  The axiom \tdx{if_def} asserts the
equation~$(if)$.
\begin{isabelle}
\isacommand{theory}\ If\ \isacommand{imports}\ FOL\isanewline
\isacommand{begin}\isanewline
\isacommand{constdefs}\isanewline
\ \ if\ ::\ "[o,o,o]=>o"\isanewline
\ \ \ "if(P,Q,R)\ ==\ P\&Q\ |\ \isachartilde P\&R"
\end{isabelle}
We create the file \texttt{If.thy} containing these declarations.  (This file
is on directory \texttt{FOL/ex} in the Isabelle distribution.)  Typing
\begin{isabelle}
use_thy "If";
\end{isabelle}
loads that theory and sets it to be the current context.

\subsection{Deriving the introduction rule}

The derivations of the introduction and elimination rules demonstrate the
methods for rewriting with definitions.  Classical reasoning is required,
so we use \isa{blast}.

The introduction rule, given the premises $P\Imp Q$ and $\neg P\Imp R$,
concludes $if(P,Q,R)$.  We propose this lemma and immediately simplify
using \isa{if\_def} to expand the definition of \isa{if} in the
subgoal.
\begin{isabelle}
\isacommand{lemma}\ ifI: "[|\ P\ ==>\ Q;\ \isachartilde P\ ==>\ R\
|]\ ==>\ if(P,Q,R)"\isanewline
\ 1.\ \isasymlbrakk P\ \isasymLongrightarrow \ Q;\ \isasymnot \ P\ \isasymLongrightarrow \ R\isasymrbrakk \ \isasymLongrightarrow \ if(P,\ Q,\ R)
\isanewline
\ 1.\ \isasymlbrakk P\ \isasymLongrightarrow \ Q;\ \isasymnot \ P\ \isasymLongrightarrow \ R\isasymrbrakk \ \isasymLongrightarrow \ P\ \isasymand \ Q\ \isasymor \ \isasymnot \ P\ \isasymand \
R
\end{isabelle}
The rule's premises, although expressed using meta-level implication
(\isa{\isasymLongrightarrow}) are passed as ordinary implications to
\methdx{blast}.
\begin{isabelle}
\isacommand{apply}\ blast\isanewline
No\ subgoals!\isanewline
\isacommand{done}
\end{isabelle}

\subsection{Deriving the elimination rule}
The elimination rule has three premises, two of which are themselves rules.
The conclusion is simply $S$.
\begin{isabelle}
\isacommand{lemma}\ ifE:\isanewline
\ \ \ "[|\ if(P,Q,R);\ \ [|P;\ Q|]\ ==>\ S;\ [|\isachartilde P;\ R|]\ ==>\ S\ |]\ ==>\ S"\isanewline
\ 1.\ \isasymlbrakk if(P,\ Q,\ R);\ \isasymlbrakk P;\ Q\isasymrbrakk \ \isasymLongrightarrow \ S;\ \isasymlbrakk \isasymnot \ P;\ R\isasymrbrakk \ \isasymLongrightarrow \ S\isasymrbrakk \ \isasymLongrightarrow \ S%
\isanewline
\ 1.\ \isasymlbrakk P\ \isasymand \ Q\ \isasymor \ \isasymnot \ P\ \isasymand \ R;\ \isasymlbrakk P;\ Q\isasymrbrakk \ \isasymLongrightarrow \ S;\ \isasymlbrakk \isasymnot \ P;\ R\isasymrbrakk \ \isasymLongrightarrow \ S\isasymrbrakk \ \isasymLongrightarrow \ S%
\end{isabelle}
The proof script is the same as before: \isa{simp} followed by
\isa{blast}:
\begin{isabelle}
\isacommand{apply}\ blast\isanewline
No\ subgoals!\isanewline
\isacommand{done}
\end{isabelle}

\subsection{Using the derived rules}
Our new derived rules, \tdx{ifI} and~\tdx{ifE}, permit natural
proofs of theorems such as the following:
\begin{eqnarray*}
if(P, if(Q,A,B), if(Q,C,D)) & \bimp & if(Q,if(P,A,C),if(P,B,D)) \\
if(if(P,Q,R), A, B)         & \bimp & if(P,if(Q,A,B),if(R,A,B))
\end{eqnarray*}
Proofs also require the classical reasoning rules and the $\bimp$
introduction rule (called~\tdx{iffI}: do not confuse with~\isa{ifI}).

To display the $if$-rules in action, let us analyse a proof step by step.
\begin{isabelle}
\isacommand{lemma}\ if\_commute:\isanewline
\ \ \ \ \ "if(P,\ if(Q,A,B),\
if(Q,C,D))\ <->\ if(Q,\ if(P,A,C),\ if(P,B,D))"\isanewline
\isacommand{apply}\ (rule\ iffI)\isanewline
\ 1.\ if(P,\ if(Q,\ A,\ B),\ if(Q,\ C,\ D))\ \isasymLongrightarrow \isanewline
\isaindent{\ 1.\ }if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 2.\ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\ \isasymLongrightarrow \isanewline
\isaindent{\ 2.\ }if(P,\ if(Q,\ A,\ B),\ if(Q,\ C,\ D))
\end{isabelle}
The $if$-elimination rule can be applied twice in succession.
\begin{isabelle}
\isacommand{apply}\ (erule\ ifE)\isanewline
\ 1.\ \isasymlbrakk P;\ if(Q,\ A,\ B)\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 2.\ \isasymlbrakk \isasymnot \ P;\ if(Q,\ C,\ D)\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 3.\ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\ \isasymLongrightarrow \isanewline
\isaindent{\ 3.\ }if(P,\ if(Q,\ A,\ B),\ if(Q,\ C,\ D))
\isanewline
\isacommand{apply}\ (erule\ ifE)\isanewline
\ 1.\ \isasymlbrakk P;\ Q;\ A\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 2.\ \isasymlbrakk P;\ \isasymnot \ Q;\ B\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 3.\ \isasymlbrakk \isasymnot \ P;\ if(Q,\ C,\ D)\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 4.\ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\ \isasymLongrightarrow \isanewline
\isaindent{\ 4.\ }if(P,\ if(Q,\ A,\ B),\ if(Q,\ C,\ D))
\end{isabelle}
%
In the first two subgoals, all assumptions have been reduced to atoms.  Now
$if$-introduction can be applied.  Observe how the $if$-rules break down
occurrences of $if$ when they become the outermost connective.
\begin{isabelle}
\isacommand{apply}\ (rule\ ifI)\isanewline
\ 1.\ \isasymlbrakk P;\ Q;\ A;\ Q\isasymrbrakk \ \isasymLongrightarrow \ if(P,\ A,\ C)\isanewline
\ 2.\ \isasymlbrakk P;\ Q;\ A;\ \isasymnot \ Q\isasymrbrakk \ \isasymLongrightarrow \ if(P,\ B,\ D)\isanewline
\ 3.\ \isasymlbrakk P;\ \isasymnot \ Q;\ B\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 4.\ \isasymlbrakk \isasymnot \ P;\ if(Q,\ C,\ D)\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 5.\ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\ \isasymLongrightarrow \isanewline
\isaindent{\ 5.\ }if(P,\ if(Q,\ A,\ B),\ if(Q,\ C,\ D))
\isanewline
\isacommand{apply}\ (rule\ ifI)\isanewline
\ 1.\ \isasymlbrakk P;\ Q;\ A;\ Q;\ P\isasymrbrakk \ \isasymLongrightarrow \ A\isanewline
\ 2.\ \isasymlbrakk P;\ Q;\ A;\ Q;\ \isasymnot \ P\isasymrbrakk \ \isasymLongrightarrow \ C\isanewline
\ 3.\ \isasymlbrakk P;\ Q;\ A;\ \isasymnot \ Q\isasymrbrakk \ \isasymLongrightarrow \ if(P,\ B,\ D)\isanewline
\ 4.\ \isasymlbrakk P;\ \isasymnot \ Q;\ B\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 5.\ \isasymlbrakk \isasymnot \ P;\ if(Q,\ C,\ D)\isasymrbrakk \ \isasymLongrightarrow \ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\isanewline
\ 6.\ if(Q,\ if(P,\ A,\ C),\ if(P,\ B,\ D))\ \isasymLongrightarrow \isanewline
\isaindent{\ 6.\ }if(P,\ if(Q,\ A,\ B),\ if(Q,\ C,\ D))
\end{isabelle}
Where do we stand?  The first subgoal holds by assumption; the second and
third, by contradiction.  This is getting tedious.  We could use the classical
reasoner, but first let us extend the default claset with the derived rules
for~$if$.
\begin{isabelle}
\isacommand{declare}\ ifI\ [intro!]\isanewline
\isacommand{declare}\ ifE\ [elim!]
\end{isabelle}
With these declarations, we could have proved this theorem with a single
call to \isa{blast}.  Here is another example:
\begin{isabelle}
\isacommand{lemma}\ "if(if(P,Q,R),\ A,\ B)\ <->\ if(P,\ if(Q,A,B),\ if(R,A,B))"\isanewline
\ 1.\ if(if(P,\ Q,\ R),\ A,\ B)\ \isasymlongleftrightarrow \ if(P,\ if(Q,\ A,\ B),\ if(R,\ A,\ B))
\isanewline
\isacommand{by}\ blast
\end{isabelle}

\subsection{Derived rules versus definitions}
Dispensing with the derived rules, we can treat $if$ as an
abbreviation, and let \ttindex{blast_tac} prove the expanded formula.  Let
us redo the previous proof:
\begin{isabelle}
\isacommand{lemma}\ "if(if(P,Q,R),\ A,\ B)\ <->\ if(P,\ if(Q,A,B),\ if(R,A,B))"\isanewline
\ 1.\ if(if(P,\ Q,\ R),\ A,\ B)\ \isasymlongleftrightarrow \ if(P,\ if(Q,\ A,\ B),\ if(R,\ A,\ B))
\end{isabelle}
This time, we simply unfold using the definition of $if$:
\begin{isabelle}
\ 1.\ (P\ \isasymand \ Q\ \isasymor \ \isasymnot \ P\ \isasymand \ R)\ \isasymand \ A\ \isasymor \ (\isasymnot \ P\ \isasymor \ \isasymnot \ Q)\ \isasymand \ (P\ \isasymor \ \isasymnot \ R)\ \isasymand \ B\ \isasymlongleftrightarrow \isanewline
\isaindent{\ 1.\ }P\ \isasymand \ (Q\ \isasymand \ A\ \isasymor \ \isasymnot \ Q\ \isasymand \ B)\ \isasymor \ \isasymnot \ P\ \isasymand \ (R\ \isasymand \ A\ \isasymor \ \isasymnot \ R\ \isasymand \ B)
\end{isabelle}
We are left with a subgoal in pure first-order logic, and it falls to
\isa{blast}:
\begin{isabelle}
\isacommand{apply}\ blast\isanewline
No\ subgoals!
\end{isabelle}
Expanding definitions reduces the extended logic to the base logic.  This
approach has its merits, but it can be slow.  In these examples, proofs
using the derived rules for~\isa{if} run about six
times faster  than proofs using just the rules of first-order logic.

Expanding definitions can also make it harder to diagnose errors.
Suppose we are having difficulties in proving some goal.  If by expanding
definitions we have made it unreadable, then we have little hope of
diagnosing the problem.

Attempts at program verification often yield invalid assertions.
Let us try to prove one:
\begin{isabelle}
\isacommand{lemma}\ "if(if(P,Q,R),\ A,\ B)\ <->\ if(P,\ if(Q,A,B),\ if(R,B,A))"\isanewline
\ 1.\ if(if(P,\ Q,\ R),\ A,\ B)\ \isasymlongleftrightarrow \ if(P,\ if(Q,\ A,\ B),\ if(R,\ B,\
A))
\end{isabelle}
Calling \isa{blast} yields an uninformative failure message. We can
get a closer look at the situation by applying \methdx{auto}.
\begin{isabelle}
\isacommand{apply}\ auto\isanewline
\ 1.\ \isasymlbrakk A;\ \isasymnot \ P;\ R\isasymrbrakk \ \isasymLongrightarrow \ B\isanewline
\ 2.\ \isasymlbrakk B;\ \isasymnot \ P;\ \isasymnot \ R\isasymrbrakk \ \isasymLongrightarrow \ A\isanewline
\ 3.\ \isasymlbrakk B;\ \isasymnot \ P;\ R\isasymrbrakk \ \isasymLongrightarrow \ A\isanewline
\ 4.\ \isasymlbrakk \isasymnot \ R;\ A;\ \isasymnot \ B;\ \isasymnot \ P\isasymrbrakk \ \isasymLongrightarrow \
False
\end{isabelle}
Subgoal~1 is unprovable and yields a countermodel: $P$ and~$B$ are false
while~$R$ and~$A$ are true.  This truth assignment reduces the main goal to
$true\bimp false$, which is of course invalid.

We can repeat this analysis by expanding definitions, using just the rules of
first-order logic:
\begin{isabelle}
\isacommand{lemma}\ "if(if(P,Q,R),\ A,\ B)\ <->\ if(P,\ if(Q,A,B),\ if(R,B,A))"\isanewline
\ 1.\ if(if(P,\ Q,\ R),\ A,\ B)\ \isasymlongleftrightarrow \ if(P,\ if(Q,\ A,\ B),\ if(R,\ B,\
A))
\isanewline
\ 1.\ (P\ \isasymand \ Q\ \isasymor \ \isasymnot \ P\ \isasymand \ R)\ \isasymand \ A\ \isasymor \ (\isasymnot \ P\ \isasymor \ \isasymnot \ Q)\ \isasymand \ (P\ \isasymor \ \isasymnot \ R)\ \isasymand \ B\ \isasymlongleftrightarrow \isanewline
\isaindent{\ 1.\ }P\ \isasymand \ (Q\ \isasymand \ A\ \isasymor \ \isasymnot \ Q\ \isasymand \ B)\ \isasymor \ \isasymnot \ P\ \isasymand \ (R\ \isasymand \ B\ \isasymor \ \isasymnot \ R\ \isasymand \ A)
\end{isabelle}
Again \isa{blast} would fail, so we try \methdx{auto}:
\begin{isabelle}
\isacommand{apply}\ (auto)\ \isanewline
\ 1.\ \isasymlbrakk A;\ \isasymnot \ P;\ R\isasymrbrakk \ \isasymLongrightarrow \ B\isanewline
\ 2.\ \isasymlbrakk A;\ \isasymnot \ P;\ R;\ \isasymnot \ B\isasymrbrakk \ \isasymLongrightarrow \ Q\isanewline
\ 3.\ \isasymlbrakk B;\ \isasymnot \ R;\ \isasymnot \ P;\ \isasymnot \ A\isasymrbrakk \ \isasymLongrightarrow \ False\isanewline
\ 4.\ \isasymlbrakk B;\ \isasymnot \ P;\ \isasymnot \ A;\ \isasymnot \ R;\ Q\isasymrbrakk \ \isasymLongrightarrow \ False\isanewline
\ 5.\ \isasymlbrakk B;\ \isasymnot \ Q;\ \isasymnot \ R;\ \isasymnot \ P;\ \isasymnot \ A\isasymrbrakk \ \isasymLongrightarrow \ False\isanewline
\ 6.\ \isasymlbrakk B;\ \isasymnot \ A;\ \isasymnot \ P;\ R\isasymrbrakk \ \isasymLongrightarrow \ False\isanewline
\ 7.\ \isasymlbrakk \isasymnot \ P;\ A;\ \isasymnot \ B;\ \isasymnot \ R\isasymrbrakk \ \isasymLongrightarrow \ False\isanewline
\ 8.\ \isasymlbrakk \isasymnot \ P;\ A;\ \isasymnot \ B;\ \isasymnot \ R\isasymrbrakk \ \isasymLongrightarrow \ Q%
\end{isabelle}
Subgoal~1 yields the same countermodel as before.  But each proof step has
taken six times as long, and the final result contains twice as many subgoals.

Expanding your definitions usually makes proofs more difficult.  This is
why the classical prover has been designed to accept derived rules.

\index{first-order logic|)}