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doc-src/TutorialI/Misc/AdvancedInd.thy

author | nipkow |

Tue Aug 29 16:05:13 2000 +0200 (2000-08-29) | |

changeset 9723 | a977245dfc8a |

parent 9689 | 751fde5307e4 |

child 9792 | bbefb6ce5cb2 |

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

*** empty log message ***

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 \isa{hd} and \isa{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 \isa{hd} and

39 \isa{last} return when applied to \isa{[]}.

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

42 formula is only the conclusion, the occurrence of \isa{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 \isa{\isasymlongrightarrow}.}

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 \isa{auto} finishes the whole proof.

64 If \isa{hd\_rev} is meant to be a simplification rule, you are done. But if you

65 really need the \isa{\isasymLongrightarrow}-version of \isa{hd\_rev}, for

66 example because you want to apply it as an introduction rule, you need to

67 derive it separately, by combining it with modus ponens:

68 *};

70 lemmas hd_revI = hd_rev[THEN mp];

72 text{*\noindent

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

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

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

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

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

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

80 which can yield a fairly complex conclusion.

81 Here is a simple example (which is proved by \isa{blast}):

82 *};

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

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

87 text{*\noindent

88 You can get the desired lemma by explicit

89 application of modus ponens and \isa{spec}:

90 *};

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

94 text{*\noindent

95 or the wholesale stripping of \isa{\isasymforall} and

96 \isa{\isasymlongrightarrow} in the conclusion via \isa{rulify}

97 *};

99 lemmas myrule = simple[rulify];

101 text{*\noindent

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

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

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

105 *};

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

108 (*<*)

109 by blast;

110 (*>*)

112 text{*

113 \bigskip

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

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

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

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

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

120 afterwards. An example appears below.

121 *};

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

125 text{*

126 So far, inductive proofs where by structural induction for

127 primitive recursive functions and recursion induction for total recursive

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

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

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

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

133 Structural induction on \isa{nat} is

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

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

136 holds for all $m<n$. In Isabelle, this is the theorem \isa{less\_induct}:

137 \begin{quote}

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

139 \end{quote}

140 Here is an example of its application.

141 *};

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

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

146 text{*\noindent

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

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

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

150 satisfies it.}

151 for \isa{f} it follows that @{term"n <= f n"}, which can

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

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

154 *};

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

158 txt{*\noindent

159 To perform induction on \isa{k} using \isa{less\_induct}, we use the same

160 general induction method as for recursion induction (see

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

162 *};

164 apply(induct_tac k rule:less_induct);

165 (*<*)

166 apply(rule allI);

167 apply(case_tac i);

168 apply(simp);

169 (*>*)

170 txt{*\noindent

171 which leaves us with the following proof state:

172 \begin{isabelle}

173 \ 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

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

175 \end{isabelle}

176 After stripping the \isa{\isasymforall i}, the proof continues with a case

177 distinction on \isa{i}. The case \isa{i = 0} is trivial and we focus on the

178 other case:

179 \begin{isabelle}

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

181 \ \ \ \ \ \ \ {\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

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

183 \end{isabelle}

184 *};

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

188 text{*\noindent

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

190 The proof goes like this (writing \isa{j} instead of \isa{nat}).

191 Since @{term"i = Suc j"} it suffices to show

192 @{term"j < f(Suc j)"} (by \isa{Suc\_leI}: @{thm"Suc_leI"[no_vars]}). This is

193 proved as follows. From \isa{f\_ax} we have @{term"f (f j) < f (Suc j)"}

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

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

196 (\isa{le_less_trans}: @{thm"le_less_trans"[no_vars]}).

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

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

199 \isa{le_less_trans}).

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

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

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

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

205 proofs.

207 We can now derive the desired @{term"i <= f i"} from \isa{f\_incr}:

208 *};

210 lemmas f_incr = f_incr_lem[rulify, OF refl];

212 text{*\noindent

213 The final \isa{refl} gets rid of the premise \isa{?k = f ?i}. Again, we could

214 have included this derivation in the original statement of the lemma:

215 *};

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

218 (*<*)oops;(*>*)

220 text{*

221 \begin{exercise}

222 From the above axiom and lemma for \isa{f} show that \isa{f} is the identity.

223 \end{exercise}

225 In general, \isa{induct\_tac} can be applied with any rule \isa{r}

226 whose conclusion is of the form \isa{?P ?x1 \dots ?xn}, in which case the

227 format is

228 \begin{ttbox}

229 apply(induct_tac y1 ... yn rule: r)

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

231 where \isa{y1}, \dots, \isa{yn} are variables in the first subgoal.

232 In fact, \isa{induct\_tac} even allows the conclusion of

233 \isa{r} to be an (iterated) conjunction of formulae of the above form, in

234 which case the application is

235 \begin{ttbox}

236 apply(induct_tac y1 ... yn and ... and z1 ... zm rule: r)

237 \end{ttbox}

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 \isa{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) ==> \\<forall>m<n. P m";

251 apply(induct_tac n);

253 txt{*\noindent

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

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

256 hypothesis and @{term"m = n"} follow from the assumption again using

257 the induction hypothesis:

258 *};

260 apply(blast);

261 (* apply(blast elim:less_SucE); *)

262 ML"set quick_and_dirty"

263 sorry;

264 ML"reset quick_and_dirty"

266 text{*\noindent

267 The elimination rule \isa{less_SucE} expresses the case distinction:

268 \begin{quote}

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

270 \end{quote}

272 Now it is straightforward to derive the original version of

273 \isa{less\_induct} by manipulting the conclusion of the above lemma:

274 instantiate \isa{n} by @{term"Suc n"} and \isa{m} by \isa{n} and

275 remove the trivial condition @{term"n < Sc n"}. Fortunately, this

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

277 desired goal:

278 *};

280 theorem less_induct: "(\\<And>n::nat. \\<forall>m<n. P m \\<Longrightarrow> P n) ==> P n";

281 by(insert induct_lem, blast);

283 text{*\noindent

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

285 inductions, \emph{wellfounded induction} (\isa{wf\_induct}):

286 \begin{quote}

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

288 \end{quote}

289 where @{term"wf r"} means that the relation \isa{r} is wellfounded.

290 For example \isa{less\_induct} is the special case where \isa{r} is \isa{<} on @{typ"nat"}.

291 For details see the library.

292 *};

294 (*<*)

295 end

296 (*>*)