summary |
shortlog |
changelog |
graph |
tags |
branches |
files |
changeset |
file |
revisions |
annotate |
diff |
raw

doc-src/TutorialI/Misc/AdvancedInd.thy

author | wenzelm |

Tue Sep 12 22:13:23 2000 +0200 (2000-09-12) | |

changeset 9941 | fe05af7ec816 |

parent 9933 | 9feb1e0c4cb3 |

child 10186 | 499637e8f2c6 |

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

renamed atts: rulify to rule_format, elimify to elim_format;

1 (*<*)

2 theory AdvancedInd = Main:;

3 (*>*)

5 text{*\noindent

6 Now that we have learned about rules and logic, we take another look at the

7 finer points of induction. The two questions we answer are: what to do if the

8 proposition to be proved is not directly amenable to induction, and how to

9 utilize and even derive new induction schemas.

10 *};

12 subsection{*Massaging the proposition\label{sec:ind-var-in-prems}*};

14 text{*

15 \noindent

16 So far we have assumed that the theorem we want to prove is already in a form

17 that is amenable to induction, but this is not always the case:

18 *};

20 lemma "xs \<noteq> [] \<Longrightarrow> hd(rev xs) = last xs";

21 apply(induct_tac xs);

23 txt{*\noindent

24 (where @{term"hd"} and @{term"last"} return the first and last element of a

25 non-empty list)

26 produces the warning

27 \begin{quote}\tt

28 Induction variable occurs also among premises!

29 \end{quote}

30 and leads to the base case

31 \begin{isabelle}

32 \ 1.\ xs\ {\isasymnoteq}\ []\ {\isasymLongrightarrow}\ hd\ (rev\ [])\ =\ last\ []

33 \end{isabelle}

34 which, after simplification, becomes

35 \begin{isabelle}

36 \ 1.\ xs\ {\isasymnoteq}\ []\ {\isasymLongrightarrow}\ hd\ []\ =\ last\ []

37 \end{isabelle}

38 We cannot prove this equality because we do not know what @{term"hd"} and

39 @{term"last"} return when applied to @{term"[]"}.

41 The point is that we have violated the above warning. Because the induction

42 formula is only the conclusion, the occurrence of @{term"xs"} in the premises is

43 not modified by induction. Thus the case that should have been trivial

44 becomes unprovable. Fortunately, the solution is easy:

45 \begin{quote}

46 \emph{Pull all occurrences of the induction variable into the conclusion

47 using @{text"\<longrightarrow>"}.}

48 \end{quote}

49 This means we should prove

50 *};

51 (*<*)oops;(*>*)

52 lemma hd_rev: "xs \<noteq> [] \<longrightarrow> hd(rev xs) = last xs";

53 (*<*)

54 by(induct_tac xs, auto);

55 (*>*)

57 text{*\noindent

58 This time, induction leaves us with the following base case

59 \begin{isabelle}

60 \ 1.\ []\ {\isasymnoteq}\ []\ {\isasymlongrightarrow}\ hd\ (rev\ [])\ =\ last\ []

61 \end{isabelle}

62 which is trivial, and @{text"auto"} finishes the whole proof.

64 If @{thm[source]hd_rev} is meant to be a simplification rule, you are

65 done. But if you really need the @{text"\<Longrightarrow>"}-version of

66 @{thm[source]hd_rev}, for example because you want to apply it as an

67 introduction rule, you need to derive it separately, by combining it with

68 modus ponens:

69 *};

71 lemmas hd_revI = hd_rev[THEN mp];

73 text{*\noindent

74 which yields the lemma we originally set out to prove.

76 In case there are multiple premises $A@1$, \dots, $A@n$ containing the

77 induction variable, you should turn the conclusion $C$ into

78 \[ A@1 \longrightarrow \cdots A@n \longrightarrow C \]

79 (see the remark?? in \S\ref{??}).

80 Additionally, you may also have to universally quantify some other variables,

81 which can yield a fairly complex conclusion.

82 Here is a simple example (which is proved by @{text"blast"}):

83 *};

85 lemma simple: "\<forall>y. A y \<longrightarrow> B y \<longrightarrow> B y & A y";

86 (*<*)by blast;(*>*)

88 text{*\noindent

89 You can get the desired lemma by explicit

90 application of modus ponens and @{thm[source]spec}:

91 *};

93 lemmas myrule = simple[THEN spec, THEN mp, THEN mp];

95 text{*\noindent

96 or the wholesale stripping of @{text"\<forall>"} and

97 @{text"\<longrightarrow>"} in the conclusion via @{text"rule_format"}

98 *};

100 lemmas myrule = simple[rule_format];

102 text{*\noindent

103 yielding @{thm"myrule"[no_vars]}.

104 You can go one step further and include these derivations already in the

105 statement of your original lemma, thus avoiding the intermediate step:

106 *};

108 lemma myrule[rule_format]: "\<forall>y. A y \<longrightarrow> B y \<longrightarrow> B y & A y";

109 (*<*)

110 by blast;

111 (*>*)

113 text{*

114 \bigskip

116 A second reason why your proposition may not be amenable to induction is that

117 you want to induct on a whole term, rather than an individual variable. In

118 general, when inducting on some term $t$ you must rephrase the conclusion as

119 \[ \forall y@1 \dots y@n.~ x = t \longrightarrow C \] where $y@1 \dots y@n$

120 are the free variables in $t$ and $x$ is new, and perform induction on $x$

121 afterwards. An example appears below.

122 *};

124 subsection{*Beyond structural and recursion induction*};

126 text{*

127 So far, inductive proofs where by structural induction for

128 primitive recursive functions and recursion induction for total recursive

129 functions. But sometimes structural induction is awkward and there is no

130 recursive function in sight either that could furnish a more appropriate

131 induction schema. In such cases some existing standard induction schema can

132 be helpful. We show how to apply such induction schemas by an example.

134 Structural induction on @{typ"nat"} is

135 usually known as ``mathematical induction''. There is also ``complete

136 induction'', where you must prove $P(n)$ under the assumption that $P(m)$

137 holds for all $m<n$. In Isabelle, this is the theorem @{thm[source]nat_less_induct}:

138 @{thm[display]"nat_less_induct"[no_vars]}

139 Here is an example of its application.

140 *};

142 consts f :: "nat => nat";

143 axioms f_ax: "f(f(n)) < f(Suc(n))";

145 text{*\noindent

146 From the above axiom\footnote{In general, the use of axioms is strongly

147 discouraged, because of the danger of inconsistencies. The above axiom does

148 not introduce an inconsistency because, for example, the identity function

149 satisfies it.}

150 for @{term"f"} it follows that @{prop"n <= f n"}, which can

151 be proved by induction on @{term"f n"}. Following the recipy outlined

152 above, we have to phrase the proposition as follows to allow induction:

153 *};

155 lemma f_incr_lem: "\<forall>i. k = f i \<longrightarrow> i \<le> f i";

157 txt{*\noindent

158 To perform induction on @{term"k"} using @{thm[source]nat_less_induct}, we use the same

159 general induction method as for recursion induction (see

160 \S\ref{sec:recdef-induction}):

161 *};

163 apply(induct_tac k rule: nat_less_induct);

164 (*<*)

165 apply(rule allI);

166 apply(case_tac i);

167 apply(simp);

168 (*>*)

169 txt{*\noindent

170 which leaves us with the following proof state:

171 \begin{isabelle}

172 \ 1.\ {\isasymAnd}\mbox{n}.\ {\isasymforall}\mbox{m}.\ \mbox{m}\ <\ \mbox{n}\ {\isasymlongrightarrow}\ ({\isasymforall}\mbox{i}.\ \mbox{m}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i})\isanewline

173 \ \ \ \ \ \ \ {\isasymLongrightarrow}\ {\isasymforall}\mbox{i}.\ \mbox{n}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i}

174 \end{isabelle}

175 After stripping the @{text"\<forall>i"}, the proof continues with a case

176 distinction on @{term"i"}. The case @{prop"i = 0"} is trivial and we focus on

177 the other case:

178 \begin{isabelle}

179 \ 1.\ {\isasymAnd}\mbox{n}\ \mbox{i}\ \mbox{nat}.\isanewline

180 \ \ \ \ \ \ \ {\isasymlbrakk}{\isasymforall}\mbox{m}.\ \mbox{m}\ <\ \mbox{n}\ {\isasymlongrightarrow}\ ({\isasymforall}\mbox{i}.\ \mbox{m}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i});\ \mbox{i}\ =\ Suc\ \mbox{nat}{\isasymrbrakk}\isanewline

181 \ \ \ \ \ \ \ {\isasymLongrightarrow}\ \mbox{n}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i}

182 \end{isabelle}

183 *};

185 by(blast intro!: f_ax Suc_leI intro: le_less_trans);

187 text{*\noindent

188 It is not surprising if you find the last step puzzling.

189 The proof goes like this (writing @{term"j"} instead of @{typ"nat"}).

190 Since @{prop"i = Suc j"} it suffices to show

191 @{prop"j < f(Suc j)"} (by @{thm[source]Suc_leI}: @{thm"Suc_leI"[no_vars]}). This is

192 proved as follows. From @{thm[source]f_ax} we have @{prop"f (f j) < f (Suc j)"}

193 (1) which implies @{prop"f j <= f (f j)"} (by the induction hypothesis).

194 Using (1) once more we obtain @{prop"f j < f(Suc j)"} (2) by transitivity

195 (@{thm[source]le_less_trans}: @{thm"le_less_trans"[no_vars]}).

196 Using the induction hypothesis once more we obtain @{prop"j <= f j"}

197 which, together with (2) yields @{prop"j < f (Suc j)"} (again by

198 @{thm[source]le_less_trans}).

200 This last step shows both the power and the danger of automatic proofs: they

201 will usually not tell you how the proof goes, because it can be very hard to

202 translate the internal proof into a human-readable format. Therefore

203 \S\ref{sec:part2?} introduces a language for writing readable yet concise

204 proofs.

206 We can now derive the desired @{prop"i <= f i"} from @{text"f_incr"}:

207 *};

209 lemmas f_incr = f_incr_lem[rule_format, OF refl];

211 text{*\noindent

212 The final @{thm[source]refl} gets rid of the premise @{text"?k = f ?i"}. Again,

213 we could have included this derivation in the original statement of the lemma:

214 *};

216 lemma f_incr[rule_format, OF refl]: "\<forall>i. k = f i \<longrightarrow> i \<le> f i";

217 (*<*)oops;(*>*)

219 text{*

220 \begin{exercise}

221 From the above axiom and lemma for @{term"f"} show that @{term"f"} is the

222 identity.

223 \end{exercise}

225 In general, @{text"induct_tac"} can be applied with any rule $r$

226 whose conclusion is of the form ${?}P~?x@1 \dots ?x@n$, in which case the

227 format is

228 \begin{quote}

229 \isacommand{apply}@{text"(induct_tac"} $y@1 \dots y@n$ @{text"rule:"} $r$@{text")"}

230 \end{quote}\index{*induct_tac}%

231 where $y@1, \dots, y@n$ are variables in the first subgoal.

232 In fact, @{text"induct_tac"} even allows the conclusion of

233 $r$ to be an (iterated) conjunction of formulae of the above form, in

234 which case the application is

235 \begin{quote}

236 \isacommand{apply}@{text"(induct_tac"} $y@1 \dots y@n$ @{text"and"} \dots\ @{text"and"} $z@1 \dots z@m$ @{text"rule:"} $r$@{text")"}

237 \end{quote}

238 *};

240 subsection{*Derivation of new induction schemas*};

242 text{*\label{sec:derive-ind}

243 Induction schemas are ordinary theorems and you can derive new ones

244 whenever you wish. This section shows you how to, using the example

245 of @{thm[source]nat_less_induct}. Assume we only have structural induction

246 available for @{typ"nat"} and want to derive complete induction. This

247 requires us to generalize the statement first:

248 *};

250 lemma induct_lem: "(\<And>n::nat. \<forall>m<n. P m \<Longrightarrow> P n) \<Longrightarrow> \<forall>m<n. P m";

251 apply(induct_tac n);

253 txt{*\noindent

254 The base case is trivially true. For the induction step (@{prop"m <

255 Suc n"}) we distinguish two cases: case @{prop"m < n"} is true by induction

256 hypothesis and case @{prop"m = n"} follows from the assumption, again using

257 the induction hypothesis:

258 *};

259 apply(blast);

260 by(blast elim:less_SucE)

262 text{*\noindent

263 The elimination rule @{thm[source]less_SucE} expresses the case distinction:

264 @{thm[display]"less_SucE"[no_vars]}

266 Now it is straightforward to derive the original version of

267 @{thm[source]nat_less_induct} by manipulting the conclusion of the above lemma:

268 instantiate @{term"n"} by @{term"Suc n"} and @{term"m"} by @{term"n"} and

269 remove the trivial condition @{prop"n < Sc n"}. Fortunately, this

270 happens automatically when we add the lemma as a new premise to the

271 desired goal:

272 *};

274 theorem nat_less_induct: "(\<And>n::nat. \<forall>m<n. P m \<Longrightarrow> P n) \<Longrightarrow> P n";

275 by(insert induct_lem, blast);

277 text{*

278 Finally we should mention that HOL already provides the mother of all

279 inductions, \emph{wellfounded induction} (@{thm[source]wf_induct}):

280 @{thm[display]"wf_induct"[no_vars]}

281 where @{term"wf r"} means that the relation @{term"r"} is wellfounded.

282 For example, theorem @{thm[source]nat_less_induct} can be viewed (and

283 derived) as a special case of @{thm[source]wf_induct} where

284 @{term"r"} is @{text"<"} on @{typ"nat"}. For details see the library.

285 *};

287 (*<*)

288 end

289 (*>*)