section\<open>Case Study: A Context Free Grammar\<close>
text\<open>\label{sec:CFG} \index{grammars!defining inductively|(}%
Grammars are nothing but shorthands forinductive definitions of nonterminals
which represent sets of strings. For example, the production
$A \to B c$ is short for \[ w \in B \Longrightarrow wc \in A \]
This section demonstrates this idea with an example
due to Hopcroft and Ullman, a grammar for generating all words with an
equal number of $a$'s and~$b$'s: \begin{eqnarray}
S &\to& \epsilon \mid b A \mid a B \nonumber\\
A &\to& a S \mid b A A \nonumber\\
B &\to& b S \mid a B B \nonumber \end{eqnarray}
At the end we say a few words about the relationship between
the original proof\<^cite>\<open>\<open>p.\ts81\<close> in HopcroftUllman\<close> and our formal version.
We start by fixing the alphabet, which consists only of \<^term>\<open>a\<close>'s and~\<^term>\<open>b\<close>'s: \<close>
datatype alfa = a | b
text\<open>\noindent For convenience we include the following easy lemmas as simplification rules: \<close>
lemma [simp]: "(x \ a) = (x = b) \ (x \ b) = (x = a)" by (case_tac x, auto)
text\<open>\noindent
Words over this alphabet are of type \<^typ>\<open>alfa list\<close>, and
the three nonterminals are declared as sets of such words.
The productions above are recast as a \emph{mutual} inductive definition\index{inductive definition!simultaneous}
of \<^term>\<open>S\<close>, \<^term>\<open>A\<close> and~\<^term>\<open>B\<close>: \<close>
inductive_set
S :: "alfa list set"and
A :: "alfa list set"and
B :: "alfa list set" where "[] \ S"
| "w \ A \ b#w \ S"
| "w \ B \ a#w \ S"
| "w \ S \ a#w \ A"
| "\ v\A; w\A \ \ b#v@w \ A"
| "w \ S \ b#w \ B"
| "\ v \ B; w \ B \ \ a#v@w \ B"
text\<open>\noindent
First we show that all words in\<^term>\<open>S\<close> contain the same number of \<^term>\<open>a\<close>'s and \<^term>\<open>b\<close>'s. Since the definition of \<^term>\<open>S\<close> is by mutual induction, so is the proof: we show at the same time that all words in \<^term>\<open>A\<close> contain one more \<^term>\<open>a\<close> than \<^term>\<open>b\<close> and all words in \<^term>\<open>B\<close> contain one more \<^term>\<open>b\<close> than \<^term>\<open>a\<close>. \<close>
lemma correctness: "(w \ S \ size[x\w. x=a] = size[x\w. x=b]) \
(w \<in> A \<longrightarrow> size[x\<leftarrow>w. x=a] = size[x\<leftarrow>w. x=b] + 1) \<and>
(w \<in> B \<longrightarrow> size[x\<leftarrow>w. x=b] = size[x\<leftarrow>w. x=a] + 1)"
txt\<open>\noindent
These propositions are expressed with the help of the predefined \<^term>\<open>filter\<close> function on lists, which has the convenient syntax \<open>[x\<leftarrow>xs. P
x]\<close>, the list of all elements \<^term>\<open>x\<close> in \<^term>\<open>xs\<close> such that \<^prop>\<open>P x\<close>
holds. Remember that on lists \<open>size\<close> and \<open>length\<close> are synonymous.
The proof itself isby rule inductionand afterwards automatic: \<close>
by (rule S_A_B.induct, auto)
text\<open>\noindent
This may seem surprising at first, andis indeed an indication of the power
of inductive definitions. But it isalso quite straightforward. For example,
consider the production $A \to b A A$: if $v,w \in A$ and the elements of $A$
contain one more $a$ than~$b$'s, then $bvw$ must again contain one more $a$
than~$b$'s.
As usual, the correctness of syntactic descriptions is easy, but completeness is hard: does \<^term>\<open>S\<close> contain \emph{all} words with an equal number of \<^term>\<open>a\<close>'s and \<^term>\<open>b\<close>'s? It turns out that this proof requires the
following lemma: every string with two more \<^term>\<open>a\<close>'s than \<^term>\<open>b\<close>'s can be cut somewhere such that each half has one more \<^term>\<open>a\<close> than \<^term>\<open>b\<close>. This is best seen by imagining counting the difference between the
number of \<^term>\<open>a\<close>'s and \<^term>\<open>b\<close>'s starting at the left end of the
word. We start with 0 andend (at the right end) with 2. Since each move to the
right increases or decreases the difference by 1, we must have passed through
1 on our way from 0 to 2. Formally, we appeal to the following discrete
intermediate valuetheorem @{thm[source]nat0_intermed_int_val}
@{thm[display,margin=60]nat0_intermed_int_val[no_vars]} where\<^term>\<open>f\<close> is of type \<^typ>\<open>nat \<Rightarrow> int\<close>, \<^typ>\<open>int\<close> are the integers, \<open>\<bar>.\<bar>\<close> is the absolute value function\footnote{See
Table~\ref{tab:ascii} in the Appendix for the correct \textsc{ascii} syntax.}, and\<^term>\<open>1::int\<close> is the integer 1 (see \S\ref{sec:numbers}).
First we show that our specific function, the difference between the
numbers of \<^term>\<open>a\<close>'s and \<^term>\<open>b\<close>'s, does indeed only change by 1 in every
move to the right. At this point we also start generalizing from\<^term>\<open>a\<close>'s and\<^term>\<open>b\<close>'s to an arbitrary property \<^term>\<open>P\<close>. Otherwise we would have to prove the desired lemma twice, once as stated above and once with the
roles of \<^term>\<open>a\<close>'s and \<^term>\<open>b\<close>'s interchanged. \<close>
lemma step1: "\i < size w. \<bar>(int(size[x\<leftarrow>take (i+1) w. P x])-int(size[x\<leftarrow>take (i+1) w. \<not>P x]))
- (int(size[x\<leftarrow>take i w. P x])-int(size[x\<leftarrow>take i w. \<not>P x]))\<bar> \<le> 1"
txt\<open>\noindent
The lemmais a bit hard to read because of the coercion function \<open>int :: nat \<Rightarrow> int\<close>. It is required because \<^term>\<open>size\<close> returns
a natural number, but subtraction on type~\<^typ>\<open>nat\<close> will do the wrong thing. Function\<^term>\<open>take\<close> is predefined and \<^term>\<open>take i xs\<close> is the prefix of
length \<^term>\<open>i\<close> of \<^term>\<open>xs\<close>; below we also need \<^term>\<open>drop i xs\<close>, which is what remains after that prefix has been dropped from\<^term>\<open>xs\<close>.
The proofisbyinduction on \<^term>\<open>w\<close>, with a trivial base case, and a not
so trivial induction step. Since it is essentially just arithmetic, we do not
discuss it. \<close>
text\<open> Finally we come to the above-mentioned lemma about cutting in half a word with two more elements of one sort than of the other sort: \<close>
lemma part1: "size[x\w. P x] = size[x\w. \P x]+2 \ \<exists>i\<le>size w. size[x\<leftarrow>take i w. P x] = size[x\<leftarrow>take i w. \<not>P x]+1"
txt\<open>\noindent
This is proved by\<open>force\<close> with the help of the intermediate value theorem,
instantiated appropriately andwith its first premise disposed of bylemma
@{thm[source]step1}: \<close>
apply(insert nat0_intermed_int_val[OF step1, of "P""w""1"]) by force
text\<open>\noindent
Lemma @{thm[source]part1} tells us only about the prefix \<^term>\<open>take i w\<close>.
An easy lemma deals with the suffix \<^term>\<open>drop i w\<close>: \<close>
lemma part2: "\size[x\take i w @ drop i w. P x] =
size[x\<leftarrow>take i w @ drop i w. \<not>P x]+2;
size[x\<leftarrow>take i w. P x] = size[x\<leftarrow>take i w. \<not>P x]+1\<rbrakk> \<Longrightarrow> size[x\<leftarrow>drop i w. P x] = size[x\<leftarrow>drop i w. \<not>P x]+1" by(simp del: append_take_drop_id)
text\<open>\noindent In the proof we have disabled the normally useful lemma \begin{isabelle}
@{thm append_take_drop_id[no_vars]} \rulename{append_take_drop_id} \end{isabelle} to allow the simplifier toapply the following lemma instead:
@{text[display]"[x\xs@ys. P x] = [x\xs. P x] @ [x\ys. P x]"}
To dispose of trivial cases automatically, the rules of the inductive definition are declared simplification rules: \<close>
declare S_A_B.intros[simp]
text\<open>\noindent
This could have been done earlier but was not necessary so far.
The completeness theorem tells us that if a word has the same number of \<^term>\<open>a\<close>'s and \<^term>\<open>b\<close>'s, then it is in \<^term>\<open>S\<close>, and similarly for\<^term>\<open>A\<close> and \<^term>\<open>B\<close>: \<close>
theorem completeness: "(size[x\w. x=a] = size[x\w. x=b] \ w \ S) \
(size[x\<leftarrow>w. x=a] = size[x\<leftarrow>w. x=b] + 1 \<longrightarrow> w \<in> A) \<and>
(size[x\<leftarrow>w. x=b] = size[x\<leftarrow>w. x=a] + 1 \<longrightarrow> w \<in> B)"
txt\<open>\noindent
The proofisbyinduction on \<^term>\<open>w\<close>. Structural induction would fail here
because, as we can see from the grammar, we need to make bigger steps than
merely appending a single letter at the front. Hence we induct on the length
of \<^term>\<open>w\<close>, using the induction rule @{thm[source]length_induct}: \<close>
apply(induct_tac w rule: length_induct) apply(rename_tac w)
txt\<open>\noindent
The \<open>rule\<close> parameter tells \<open>induct_tac\<close> explicitly which induction
rule touse. For details see \S\ref{sec:complete-ind} below. In this case the result is that we may assume the lemma already
holds for all words shorter than \<^term>\<open>w\<close>. Because the induction step renames
the induction variable we rename it backto\<open>w\<close>.
The proof continues with a case distinction on \<^term>\<open>w\<close>,
on whether \<^term>\<open>w\<close> is empty or not. \<close>
apply(case_tac w) apply(simp_all) (*<*)apply(rename_tac x v)(*>*)
txt\<open>\noindent
Simplification disposes of the base caseand leaves only a conjunction
of two step cases to be proved: if\<^prop>\<open>w = a#v\<close> and @{prop[display]"size[x\<in>v. x=a] = size[x\<in>v. x=b]+2"} then \<^prop>\<open>b#v \<in> A\<close>, and similarly for \<^prop>\<open>w = b#v\<close>.
We only consider the first casein detail.
After breaking the conjunction up into two cases, we can apply
@{thm[source]part1} to the assumption that \<^term>\<open>w\<close> contains two more \<^term>\<open>a\<close>'s than \<^term>\<open>b\<close>'s. \<close>
apply(rule conjI) apply(clarify) apply(frule part1[of "\x. x=a", simplified]) apply(clarify) txt\<open>\noindent
This yields an index \<^prop>\<open>i \<le> length v\<close> such that
@{prop[display]"length [x\take i v . x = a] = length [x\take i v . x = b] + 1"} With the help of @{thm[source]part2} it follows that
@{prop[display]"length [x\drop i v . x = a] = length [x\drop i v . x = b] + 1"} \<close>
txt\<open>\noindent
Now it is time to decompose \<^term>\<open>v\<close> in the conclusion \<^prop>\<open>b#v \<in> A\<close>
into \<^term>\<open>take i v @ drop i v\<close>, \<close>
apply(rule_tac n1=i and t=v in subst[OF append_take_drop_id])
txt\<open>\noindent
(the variables \<^term>\<open>n1\<close> and \<^term>\<open>t\<close> are the result of composing the theorems @{thm[source]subst} and @{thm[source]append_take_drop_id})
after which the appropriate rule of the grammar reduces the goal to the two subgoals \<^prop>\<open>take i v \<in> A\<close> and \<^prop>\<open>drop i v \<in> A\<close>: \<close>
apply(rule S_A_B.intros)
txt\<open>
Both subgoals follow from the induction hypothesis because both \<^term>\<open>take i
v\<close> and \<^term>\<open>drop i v\<close> are shorter than \<^term>\<open>w\<close>: \<close>
text\<open>
We conclude this section with a comparison of our proofwith
Hopcroft\index{Hopcroft, J. E.} and Ullman's\index{Ullman, J. D.} \<^cite>\<open>\<open>p.\ts81\<close> in HopcroftUllman\<close>. For a start, the textbook
grammar, for no good reason, excludes the empty word, thus complicating
matters just a little bit: they have 8 instead of our 7 productions.
More importantly, the proof itself is different: rather than
separating the two directions, they perform one induction on the
length of a word. This deprives them of the beauty of rule induction, andin the easy direction (correctness) their reasoning is more
detailed than our \<open>auto\<close>. For the hard part (completeness), they
consider just one of the cases that our \<open>simp_all\<close> disposes of
automatically. Then they conclude the proofby saying about the
remaining cases: ``We do this in a manner similar to our method of prooffor part (1); this part is left to the reader''. But this is
precisely the part that requires the intermediate valuetheoremand thusis not at all similar to the other cases (which are automatic in
Isabelle). The authors are at least cavalier about this point and may
even have overlooked the slight difficulty lurking in the omitted
cases. Such errors are found in many pen-and-paper proofs when they
are scrutinized formally.% \index{grammars!defining inductively|)} \<close>
(*<*)end(*>*)
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