Quelle Ifexpr.thy Sprache: Isabelle

(*<*)
theory Ifexpr imports Main begin
(*>*)

subsection\<open>Case Study: Boolean Expressions\<close>

text\<open>\label{sec:boolex}\index{boolean expressions example|(}
The aim of this case study is twofold: it shows how to model boolean
expressions and some algorithms for manipulating them, and it demonstrates
the constructs introduced above.
\<close>

subsubsection\<open>Modelling Boolean Expressions\<close>

text\<open>
We want to represent boolean expressions built up from variables and
constants by negation and conjunction. The following datatype serves exactly
that purpose:
\<close>

datatype boolex = Const bool | Var nat | Neg boolex
                | And boolex boolex

text\<open>\noindent
The two constants are represented by \<^term>\<open>Const True\<close> and
\<^term>\<open>Const False\<close>. Variables are represented by terms of the form
\<^term>\<open>Var n\<close>, where \<^term>\<open>n\<close> is a natural number (type \<^typ>\<open>nat\<close>).
For example, the formula $P@0 \land \neg P@1$ is represented by the term
\<^term>\<open>And (Var 0) (Neg(Var 1))\<close>.

\subsubsection{The Value of a Boolean Expression}

The value of a boolean expression depends on the value of its variables.
Hence the function \<open>value\<close> takes an additional parameter, an
\emph{environment} of type \<^typ>\<open>nat => bool\<close>, which maps variables to their
values:
\<close>

primrec "value" :: "boolex \ (nat \ bool) \ bool" where
"value (Const b) env = b" |
"value (Var x) env = env x" |
"value (Neg b) env = (\ value b env)" |
"value (And b c) env = (value b env \ value c env)"

text\<open>\noindent
\subsubsection{If-Expressions}

An alternative and often more efficient (because in a certain sense
canonical) representation are so-called \emph{If-expressions} built up
from constants (\<^term>\<open>CIF\<close>), variables (\<^term>\<open>VIF\<close>) and conditionals
(\<^term>\<open>IF\<close>):
\<close>

datatype ifex = CIF bool | VIF nat | IF ifex ifex ifex

text\<open>\noindent
The evaluation of If-expressions proceeds as for \<^typ>\<open>boolex\<close>:
\<close>

primrec valif :: "ifex \ (nat \ bool) \ bool" where
"valif (CIF b) env = b" |
"valif (VIF x) env = env x" |
"valif (IF b t e) env = (if valif b env then valif t env
                                        else valif e env)"

text\<open>
\subsubsection{Converting Boolean and If-Expressions}

The type \<^typ>\<open>boolex\<close> is close to the customary representation of logical
formulae, whereas \<^typ>\<open>ifex\<close> is designed for efficiency. It is easy to
translate from \<^typ>\<open>boolex\<close> into \<^typ>\<open>ifex\<close>:
\<close>

primrec bool2if :: "boolex \ ifex" where
"bool2if (Const b) = CIF b" |
"bool2if (Var x) = VIF x" |
"bool2if (Neg b) = IF (bool2if b) (CIF False) (CIF True)" |
"bool2if (And b c) = IF (bool2if b) (bool2if c) (CIF False)"

text\<open>\noindent
At last, we have something we can verify: that \<^term>\<open>bool2if\<close> preserves the
value of its argument:
\<close>

lemma "valif (bool2if b) env = value b env"

txt\<open>\noindent
The proof is canonical:
\<close>

apply(induct_tac b)
apply(auto)
done

text\<open>\noindent
In fact, all proofs in this case study look exactly like this. Hence we do
not show them below.

More interesting is the transformation of If-expressions into a normal form
where the first argument of \<^term>\<open>IF\<close> cannot be another \<^term>\<open>IF\<close> but
must be a constant or variable. Such a normal form can be computed by
repeatedly replacing a subterm of the form \<^term>\<open>IF (IF b x y) z u\<close> by
\<^term>\<open>IF b (IF x z u) (IF y z u)\<close>, which has the same value. The following
primitive recursive functions perform this task:
\<close>

primrec normif :: "ifex \ ifex \ ifex \ ifex" where
"normif (CIF b) t e = IF (CIF b) t e" |
"normif (VIF x) t e = IF (VIF x) t e" |
"normif (IF b t e) u f = normif b (normif t u f) (normif e u f)"

primrec norm :: "ifex \ ifex" where
"norm (CIF b) = CIF b" |
"norm (VIF x) = VIF x" |
"norm (IF b t e) = normif b (norm t) (norm e)"

text\<open>\noindent
Their interplay is tricky; we leave it to you to develop an
intuitive understanding. Fortunately, Isabelle can help us to verify that the
transformation preserves the value of the expression:
\<close>

theorem "valif (norm b) env = valif b env"(*<*)oops(*>*)

text\<open>\noindent
The proof is canonical, provided we first show the following simplification
lemma, which also helps to understand what \<^term>\<open>normif\<close> does:
\<close>

lemma [simp]:
  "\t e. valif (normif b t e) env = valif (IF b t e) env"
(*<*)
apply(induct_tac b)
by(auto)

theorem "valif (norm b) env = valif b env"
apply(induct_tac b)
by(auto)
(*>*)
text\<open>\noindent
Note that the lemma does not have a name, but is implicitly used in the proof
of the theorem shown above because of the \<open>[simp]\<close> attribute.

But how can we be sure that \<^term>\<open>norm\<close> really produces a normal form in
the above sense? We define a function that tests If-expressions for normality:
\<close>

primrec normal :: "ifex \ bool" where
"normal(CIF b) = True" |
"normal(VIF x) = True" |
"normal(IF b t e) = (normal t \ normal e \
     (case b of CIF b \<Rightarrow> True | VIF x \<Rightarrow> True | IF x y z \<Rightarrow> False))"

text\<open>\noindent
Now we prove \<^term>\<open>normal(norm b)\<close>. Of course, this requires a lemma about
normality of \<^term>\<open>normif\<close>:
\<close>

lemma [simp]: "\t e. normal(normif b t e) = (normal t \ normal e)"
(*<*)
apply(induct_tac b)
by(auto)

theorem "normal(norm b)"
apply(induct_tac b)
by(auto)
(*>*)

text\<open>\medskip
How do we come up with the required lemmas? Try to prove the main theorems
without them and study carefully what \<open>auto\<close> leaves unproved. This
can provide the clue.  The necessity of universal quantification
(\<open>\<forall>t e\<close>) in the two lemmas is explained in
\S\ref{sec:InductionHeuristics}

\begin{exercise}
  We strengthen the definition of a \<^const>\<open>normal\<close> If-expression as follows:
  the first argument of all \<^term>\<open>IF\<close>s must be a variable. Adapt the above
  development to this changed requirement. (Hint: you may need to formulate
  some of the goals as implications (\<open>\<longrightarrow>\<close>) rather than
  equalities (\<open>=\<close>).)
\end{exercise}
\index{boolean expressions example|)}
\<close>
(*<*)

primrec normif2 :: "ifex => ifex => ifex => ifex" where
"normif2 (CIF b) t e = (if b then t else e)" |
"normif2 (VIF x) t e = IF (VIF x) t e" |
"normif2 (IF b t e) u f = normif2 b (normif2 t u f) (normif2 e u f)"

primrec norm2 :: "ifex => ifex" where
"norm2 (CIF b) = CIF b" |
"norm2 (VIF x) = VIF x" |
"norm2 (IF b t e) = normif2 b (norm2 t) (norm2 e)"

primrec normal2 :: "ifex => bool" where
"normal2(CIF b) = True" |
"normal2(VIF x) = True" |
"normal2(IF b t e) = (normal2 t & normal2 e &
     (case b of CIF b => False | VIF x => True | IF x y z => False))"

lemma [simp]:
  "\t e. valif (normif2 b t e) env = valif (IF b t e) env"
apply(induct b)
by(auto)

theorem "valif (norm2 b) env = valif b env"
apply(induct b)
by(auto)

lemma [simp]: "\t e. normal2 t & normal2 e --> normal2(normif2 b t e)"
apply(induct b)
by(auto)

theorem "normal2(norm2 b)"
apply(induct b)
by(auto)

end
(*>*)

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