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--
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     1 (*<*)

     2 theory AdvancedInd = Main:;

     3 (*>*)

     4

     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 *};

    11

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

    13

    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 *};

    19

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

    21 apply(induct_tac xs);

    22

    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{[]}.

    40

    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 (*>*)

    56

    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.

    63

    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 *};

    69

    70 lemmas hd_revI = hd_rev[THEN mp];

    71

    72 text{*\noindent

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

    74

    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 *};

    83

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

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

    86

    87 text{*\noindent

    88 You can get the desired lemma by explicit

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

    90 *};

    91

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

    93

    94 text{*\noindent

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

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

    97 *};

    98

    99 lemmas myrule = simple[rulify];

   100

   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 *};

   106

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

   108 (*<*)

   109 by blast;

   110 (*>*)

   111

   112 text{*

   113 \bigskip

   114

   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 *};

   122

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

   124

   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.

   132

   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 *};

   142

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

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

   145

   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 *};

   155

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

   157

   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 *};

   163

   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 *};

   185

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

   187

   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}).

   200

   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.

   206

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

   208 *};

   209

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

   211

   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 *};

   216

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

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

   219

   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}

   224

   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 *};

   239

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

   241

   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 *};

   249

   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);

   252

   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 *};

   259

   260 apply(blast);

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

   262 ML"set quick_and_dirty"

   263 sorry;

   264 ML"reset quick_and_dirty"

   265

   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}

   271

   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 *};

   279

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

   281 by(insert induct_lem, blast);

   282

   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 *};

   293

   294 (*<*)

   295 end

   296 (*>*)