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doc-src/TutorialI/Inductive/AB.thy

author | nipkow |

Tue Oct 17 13:28:57 2000 +0200 (2000-10-17) | |

changeset 10236 | 7626cb4e1407 |

parent 10225 | b9fd52525b69 |

child 10237 | 875bf54b5d74 |

permissions | -rw-r--r-- |

*** empty log message ***

1 (*<*)theory AB = Main:(*>*)

3 section{*A context free grammar*}

5 text{*

6 Grammars are nothing but shorthands for inductive definitions of nonterminals

7 which represent sets of strings. For example, the production

8 $A \to B c$ is short for

9 \[ w \in B \Longrightarrow wc \in A \]

10 This section demonstrates this idea with a standard example

11 \cite[p.\ 81]{HopcroftUllman}, a grammar for generating all words with an

12 equal number of $a$'s and $b$'s:

13 \begin{eqnarray}

14 S &\to& \epsilon \mid b A \mid a B \nonumber\\

15 A &\to& a S \mid b A A \nonumber\\

16 B &\to& b S \mid a B B \nonumber

17 \end{eqnarray}

18 At the end we say a few words about the relationship of the formalization

19 and the text in the book~\cite[p.\ 81]{HopcroftUllman}.

21 We start by fixing the alpgabet, which consists only of @{term a}'s

22 and @{term b}'s:

23 *}

25 datatype alfa = a | b;

27 text{*\noindent

28 For convenience we includ the following easy lemmas as simplification rules:

29 *}

31 lemma [simp]: "(x \<noteq> a) = (x = b) \<and> (x \<noteq> b) = (x = a)";

32 apply(case_tac x);

33 by(auto);

35 text{*\noindent

36 Words over this alphabet are of type @{typ"alfa list"}, and

37 the three nonterminals are declare as sets of such words:

38 *}

40 consts S :: "alfa list set"

41 A :: "alfa list set"

42 B :: "alfa list set";

44 text{*\noindent

45 The above productions are recast as a \emph{simultaneous} inductive

46 definition of @{term S}, @{term A} and @{term B}:

47 *}

49 inductive S A B

50 intros

51 "[] \<in> S"

52 "w \<in> A \<Longrightarrow> b#w \<in> S"

53 "w \<in> B \<Longrightarrow> a#w \<in> S"

55 "w \<in> S \<Longrightarrow> a#w \<in> A"

56 "\<lbrakk> v\<in>A; w\<in>A \<rbrakk> \<Longrightarrow> b#v@w \<in> A"

58 "w \<in> S \<Longrightarrow> b#w \<in> B"

59 "\<lbrakk> v \<in> B; w \<in> B \<rbrakk> \<Longrightarrow> a#v@w \<in> B";

61 text{*\noindent

62 First we show that all words in @{term S} contain the same number of @{term

63 a}'s and @{term b}'s. Since the definition of @{term S} is by simultaneous

64 induction, so is this proof: we show at the same time that all words in

65 @{term A} contain one more @{term a} than @{term b} and all words in @{term

66 B} contains one more @{term b} than @{term a}.

67 *}

69 lemma correctness:

70 "(w \<in> S \<longrightarrow> size[x\<in>w. x=a] = size[x\<in>w. x=b]) \<and>

71 (w \<in> A \<longrightarrow> size[x\<in>w. x=a] = size[x\<in>w. x=b] + 1) \<and>

72 (w \<in> B \<longrightarrow> size[x\<in>w. x=b] = size[x\<in>w. x=a] + 1)"

74 txt{*\noindent

75 These propositions are expressed with the help of the predefined @{term

76 filter} function on lists, which has the convenient syntax @{term"[x\<in>xs. P

77 x]"}, the list of all elements @{term x} in @{term xs} such that @{prop"P x"}

78 holds. The length of a list is usually written @{term length}, and @{term

79 size} is merely a shorthand.

81 The proof itself is by rule induction and afterwards automatic:

82 *}

84 apply(rule S_A_B.induct);

85 by(auto);

87 text{*\noindent

88 This may seem surprising at first, and is indeed an indication of the power

89 of inductive definitions. But it is also quite straightforward. For example,

90 consider the production $A \to b A A$: if $v,w \in A$ and the elements of $A$

91 contain one more $a$ than $b$'s, then $bvw$ must again contain one more $a$

92 than $b$'s.

94 As usual, the correctness of syntactic descriptions is easy, but completeness

95 is hard: does @{term S} contain \emph{all} words with an equal number of

96 @{term a}'s and @{term b}'s? It turns out that this proof requires the

97 following little lemma: every string with two more @{term a}'s than @{term

98 b}'s can be cut somehwere such that each half has one more @{term a} than

99 @{term b}. This is best seen by imagining counting the difference between the

100 number of @{term a}'s than @{term b}'s starting at the left end of the

101 word. We start at 0 and end (at the right end) with 2. Since each move to the

102 right increases or decreases the difference by 1, we must have passed through

103 1 on our way from 0 to 2. Formally, we appeal to the following discrete

104 intermediate value theorem @{thm[source]nat0_intermed_int_val}

105 @{thm[display]nat0_intermed_int_val[no_vars]}

106 where @{term f} is of type @{typ"nat \<Rightarrow> int"}, @{typ int} are the integers,

107 @{term abs} is the absolute value function, and @{term"#1::int"} is the

108 integer 1 (see \S\ref{sec:int}).

110 First we show that the our specific function, the difference between the

111 numbers of @{term a}'s and @{term b}'s, does indeed only change by 1 in every

112 move to the right. At this point we also start generalizing from @{term a}'s

113 and @{term b}'s to an arbitrary property @{term P}. Otherwise we would have

114 to prove the desired lemma twice, once as stated above and once with the

115 roles of @{term a}'s and @{term b}'s interchanged.

116 *}

118 lemma step1: "\<forall>i < size w.

119 abs((int(size[x\<in>take (i+1) w. P x]) -

120 int(size[x\<in>take (i+1) w. \<not>P x]))

121 -

122 (int(size[x\<in>take i w. P x]) -

123 int(size[x\<in>take i w. \<not>P x]))) <= #1";

125 txt{*\noindent

126 The lemma is a bit hard to read because of the coercion function

127 @{term[source]"int::nat \<Rightarrow> int"}. It is required because @{term size} returns

128 a natural number, but @{text-} on @{typ nat} will do the wrong thing.

129 Function @{term take} is predefined and @{term"take i xs"} is the prefix of

130 length @{term i} of @{term xs}; below we als need @{term"drop i xs"}, which

131 is what remains after that prefix has been dropped from @{term xs}.

133 The proof is by induction on @{term w}, with a trivial base case, and a not

134 so trivial induction step. Since it is essentially just arithmetic, we do not

135 discuss it.

136 *}

138 apply(induct w);

139 apply(simp);

140 by(force simp add:zabs_def take_Cons split:nat.split if_splits);

142 text{*

143 Finally we come to the above mentioned lemma about cutting a word with two

144 more elements of one sort than of the other sort into two halfs:

145 *}

147 lemma part1:

148 "size[x\<in>w. P x] = size[x\<in>w. \<not>P x]+2 \<Longrightarrow>

149 \<exists>i\<le>size w. size[x\<in>take i w. P x] = size[x\<in>take i w. \<not>P x]+1";

151 txt{*\noindent

152 This is proved with the help of the intermediate value theorem, instantiated

153 appropriately and with its first premise disposed of by lemma

154 @{thm[source]step1}.

155 *}

157 apply(insert nat0_intermed_int_val[OF step1, of "P" "w" "#1"]);

158 apply simp;

159 by(simp del:int_Suc add:zdiff_eq_eq sym[OF int_Suc]);

161 text{*\noindent

162 The additional lemmas are needed to mediate between @{typ nat} and @{typ int}.

164 Lemma @{thm[source]part1} tells us only about the prefix @{term"take i w"}.

165 The suffix @{term"drop i w"} is dealt with in the following easy lemma:

166 *}

169 lemma part2:

170 "\<lbrakk>size[x\<in>take i w @ drop i w. P x] =

171 size[x\<in>take i w @ drop i w. \<not>P x]+2;

172 size[x\<in>take i w. P x] = size[x\<in>take i w. \<not>P x]+1\<rbrakk>

173 \<Longrightarrow> size[x\<in>drop i w. P x] = size[x\<in>drop i w. \<not>P x]+1";

174 by(simp del:append_take_drop_id);

176 text{*\noindent

177 Lemma @{thm[source]append_take_drop_id}, @{thm append_take_drop_id[no_vars]},

178 which is generally useful, needs to be disabled for once.

180 To dispose of trivial cases automatically, the rules of the inductive

181 definition are declared simplification rules:

182 *}

184 declare S_A_B.intros[simp];

186 text{*\noindent

187 This could have been done earlier but was not necessary so far.

189 The completeness theorem tells us that if a word has the same number of

190 @{term a}'s and @{term b}'s, then it is in @{term S}, and similarly and

191 simultaneously for @{term A} and @{term B}:

192 *}

194 theorem completeness:

195 "(size[x\<in>w. x=a] = size[x\<in>w. x=b] \<longrightarrow> w \<in> S) \<and>

196 (size[x\<in>w. x=a] = size[x\<in>w. x=b] + 1 \<longrightarrow> w \<in> A) \<and>

197 (size[x\<in>w. x=b] = size[x\<in>w. x=a] + 1 \<longrightarrow> w \<in> B)";

199 txt{*\noindent

200 The proof is by induction on @{term w}. Structural induction would fail here

201 because, as we can see from the grammar, we need to make bigger steps than

202 merely appending a single letter at the front. Hence we induct on the length

203 of @{term w}, using the induction rule @{thm[source]length_induct}:

204 *}

206 apply(induct_tac w rule: length_induct);

207 (*<*)apply(rename_tac w)(*>*)

209 txt{*\noindent

210 The @{text rule} parameter tells @{text induct_tac} explicitly which induction

211 rule to use. For details see \S\ref{sec:complete-ind} below.

212 In this case the result is that we may assume the lemma already

213 holds for all words shorter than @{term w}.

215 The proof continues with a case distinction on @{term w},

216 i.e.\ if @{term w} is empty or not.

217 *}

219 apply(case_tac w);

220 apply(simp_all);

221 (*<*)apply(rename_tac x v)(*>*)

223 txt{*\noindent

224 Simplification disposes of the base case and leaves only two step

225 cases to be proved:

226 if @{prop"w = a#v"} and @{prop"size[x\<in>v. x=a] = size[x\<in>v. x=b]+2"} then

227 @{prop"b#v \<in> A"}, and similarly for @{prop"w = b#v"}.

228 We only consider the first case in detail.

230 After breaking the conjuction up into two cases, we can apply

231 @{thm[source]part1} to the assumption that @{term w} contains two more @{term

232 a}'s than @{term b}'s.

233 *}

235 apply(rule conjI);

236 apply(clarify);

237 apply(frule part1[of "\<lambda>x. x=a", simplified]);

238 apply(erule exE);

239 apply(erule conjE);

241 txt{*\noindent

242 This yields an index @{prop"i \<le> length v"} such that

243 @{prop"length [x\<in>take i v . x = a] = length [x\<in>take i v . x = b] + 1"}.

244 With the help of @{thm[source]part1} it follows that

245 @{prop"length [x\<in>drop i v . x = a] = length [x\<in>drop i v . x = b] + 1"}.

246 *}

248 apply(drule part2[of "\<lambda>x. x=a", simplified]);

249 apply(assumption);

251 txt{*\noindent

252 Now it is time to decompose @{term v} in the conclusion @{prop"b#v \<in> A"}

253 into @{term"take i v @ drop i v"},

254 after which the appropriate rule of the grammar reduces the goal

255 to the two subgoals @{prop"take i v \<in> A"} and @{prop"drop i v \<in> A"}:

256 *}

258 apply(rule_tac n1=i and t=v in subst[OF append_take_drop_id]);

259 apply(rule S_A_B.intros);

261 txt{*\noindent

262 Both subgoals follow from the induction hypothesis because both @{term"take i

263 v"} and @{term"drop i v"} are shorter than @{term w}:

264 *}

266 apply(force simp add: min_less_iff_disj);

267 apply(force split add: nat_diff_split);

269 txt{*\noindent

270 Note that the variables @{term n1} and @{term t} referred to in the

271 substitution step above come from the derived theorem @{text"subst[OF

272 append_take_drop_id]"}.

274 The case @{prop"w = b#v"} is proved completely analogously:

275 *}

277 apply(clarify);

278 apply(frule part1[of "\<lambda>x. x=b", simplified]);

279 apply(erule exE);

280 apply(erule conjE);

281 apply(drule part2[of "\<lambda>x. x=b", simplified]);

282 apply(assumption);

283 apply(rule_tac n1=i and t=v in subst[OF append_take_drop_id]);

284 apply(rule S_A_B.intros);

285 apply(force simp add:min_less_iff_disj);

286 by(force simp add:min_less_iff_disj split add: nat_diff_split);

288 text{*

289 We conclude this section with a comparison of the above proof and the one

290 in the textbook \cite[p.\ 81]{HopcroftUllman}. For a start, the texbook

291 grammar, for no good reason, excludes the empty word, which complicates

292 matters just a little bit because we now have 8 instead of our 7 productions.

294 More importantly, the proof itself is different: rather than separating the

295 two directions, they perform one induction on the length of a word. This

296 deprives them of the beauty of rule induction and in the easy direction

297 (correctness) their reasoning is more detailed than our @{text auto}. For the

298 hard part (completeness), they consider just one of the cases that our @{text

299 simp_all} disposes of automatically. Then they conclude the proof by saying

300 about the remaining cases: ``We do this in a manner similar to our method of

301 proof for part (1); this part is left to the reader''. But this is precisely

302 the part that requires the intermediate value theorem and thus is not at all

303 similar to the other cases (which are automatic in Isabelle). We conclude

304 that the authors are at least cavalier about this point and may even have

305 overlooked the slight difficulty lurking in the omitted cases. This is not

306 atypical for pen-and-paper proofs, once analysed in detail. *}

308 (*<*)end(*>*)