doc-src/TutorialI/Ifexpr/Ifexpr.thy

A new implementation for presburger arithmetic following the one suggested in technical report Chaieb Amine and Tobias Nipkow. It is generic an smaller.
the tactic has also changed and allows the abstaction over fuction occurences whose type is nat or int.

(*<*)
theory Ifexpr = Main:;
(*>*)

subsection{*Case Study: Boolean Expressions*}

text{*\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.
*}

subsubsection{*Modelling Boolean Expressions*}

text{*
We want to represent boolean expressions built up from variables and
constants by negation and conjunction. The following datatype serves exactly
that purpose:
*}

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

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

\subsubsection{The Value of a Boolean Expression}

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

consts value :: "boolex \<Rightarrow> (nat \<Rightarrow> bool) \<Rightarrow> bool";
primrec
"value (Const b) env = b"
"value (Var x)   env = env x"
"value (Neg b)   env = (\<not> value b env)"
"value (And b c) env = (value b env \<and> value c env)";

text{*\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"CIF"}), variables (@{term"VIF"}) and conditionals
(@{term"IF"}):
*}

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

text{*\noindent
The evaluation of If-expressions proceeds as for @{typ"boolex"}:
*}

consts valif :: "ifex \<Rightarrow> (nat \<Rightarrow> bool) \<Rightarrow> bool";
primrec
"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{*
\subsubsection{Converting Boolean and If-Expressions}

The type @{typ"boolex"} is close to the customary representation of logical
formulae, whereas @{typ"ifex"} is designed for efficiency. It is easy to
translate from @{typ"boolex"} into @{typ"ifex"}:
*}

consts bool2if :: "boolex \<Rightarrow> ifex";
primrec
"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{*\noindent
At last, we have something we can verify: that @{term"bool2if"} preserves the
value of its argument:
*}

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

txt{*\noindent
The proof is canonical:
*}

apply(induct_tac b);
apply(auto);
done

text{*\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"IF"} cannot be another @{term"IF"} but
must be a constant or variable. Such a normal form can be computed by
repeatedly replacing a subterm of the form @{term"IF (IF b x y) z u"} by
@{term"IF b (IF x z u) (IF y z u)"}, which has the same value. The following
primitive recursive functions perform this task:
*}

consts normif :: "ifex \<Rightarrow> ifex \<Rightarrow> ifex \<Rightarrow> ifex";
primrec
"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)";

consts norm :: "ifex \<Rightarrow> ifex";
primrec
"norm (CIF b)    = CIF b"
"norm (VIF x)    = VIF x"
"norm (IF b t e) = normif b (norm t) (norm e)";

text{*\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:
*}

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

text{*\noindent
The proof is canonical, provided we first show the following simplification
lemma, which also helps to understand what @{term"normif"} does:
*}

lemma [simp]:
  "\<forall>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{*\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 @{text"[simp]"} attribute.

But how can we be sure that @{term"norm"} really produces a normal form in
the above sense? We define a function that tests If-expressions for normality:
*}

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

text{*\noindent
Now we prove @{term"normal(norm b)"}. Of course, this requires a lemma about
normality of @{term"normif"}:
*}

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

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

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

\begin{exercise}
  We strengthen the definition of a @{term normal} If-expression as follows:
  the first argument of all @{term IF}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 (@{text"\<longrightarrow>"}) rather than
  equalities (@{text"="}).)
\end{exercise}
\index{boolean expressions example|)}
*}
(*<*)
end
(*>*)