author | nipkow |
Wed, 11 Oct 2000 09:09:06 +0200 | |
changeset 10186 | 499637e8f2c6 |
parent 9941 | fe05af7ec816 |
child 10217 | e61e7e1eacaf |
permissions | -rw-r--r-- |
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(*<*) |
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theory AdvancedInd = Main:; |
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(*>*) |
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text{*\noindent |
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Now that we have learned about rules and logic, we take another look at the |
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finer points of induction. The two questions we answer are: what to do if the |
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proposition to be proved is not directly amenable to induction, and how to |
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utilize and even derive new induction schemas. |
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*}; |
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subsection{*Massaging the proposition\label{sec:ind-var-in-prems}*}; |
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text{* |
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\noindent |
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So far we have assumed that the theorem we want to prove is already in a form |
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that is amenable to induction, but this is not always the case: |
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*}; |
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lemma "xs \<noteq> [] \<Longrightarrow> hd(rev xs) = last xs"; |
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apply(induct_tac xs); |
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txt{*\noindent |
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(where @{term"hd"} and @{term"last"} return the first and last element of a |
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non-empty list) |
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produces the warning |
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\begin{quote}\tt |
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Induction variable occurs also among premises! |
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\end{quote} |
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and leads to the base case |
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\begin{isabelle} |
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\ 1.\ xs\ {\isasymnoteq}\ []\ {\isasymLongrightarrow}\ hd\ (rev\ [])\ =\ last\ [] |
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\end{isabelle} |
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which, after simplification, becomes |
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\begin{isabelle} |
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\ 1.\ xs\ {\isasymnoteq}\ []\ {\isasymLongrightarrow}\ hd\ []\ =\ last\ [] |
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\end{isabelle} |
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We cannot prove this equality because we do not know what @{term"hd"} and |
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@{term"last"} return when applied to @{term"[]"}. |
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The point is that we have violated the above warning. Because the induction |
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formula is only the conclusion, the occurrence of @{term"xs"} in the premises is |
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not modified by induction. Thus the case that should have been trivial |
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becomes unprovable. Fortunately, the solution is easy: |
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\begin{quote} |
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\emph{Pull all occurrences of the induction variable into the conclusion |
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using @{text"\<longrightarrow>"}.} |
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\end{quote} |
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This means we should prove |
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*}; |
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(*<*)oops;(*>*) |
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lemma hd_rev: "xs \<noteq> [] \<longrightarrow> hd(rev xs) = last xs"; |
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(*<*) |
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by(induct_tac xs, auto); |
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(*>*) |
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text{*\noindent |
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This time, induction leaves us with the following base case |
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\begin{isabelle} |
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\ 1.\ []\ {\isasymnoteq}\ []\ {\isasymlongrightarrow}\ hd\ (rev\ [])\ =\ last\ [] |
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\end{isabelle} |
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which is trivial, and @{text"auto"} finishes the whole proof. |
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If @{thm[source]hd_rev} is meant to be a simplification rule, you are |
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done. But if you really need the @{text"\<Longrightarrow>"}-version of |
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@{thm[source]hd_rev}, for example because you want to apply it as an |
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introduction rule, you need to derive it separately, by combining it with |
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modus ponens: |
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*}; |
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lemmas hd_revI = hd_rev[THEN mp]; |
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text{*\noindent |
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which yields the lemma we originally set out to prove. |
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In case there are multiple premises $A@1$, \dots, $A@n$ containing the |
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induction variable, you should turn the conclusion $C$ into |
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\[ A@1 \longrightarrow \cdots A@n \longrightarrow C \] |
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(see the remark?? in \S\ref{??}). |
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Additionally, you may also have to universally quantify some other variables, |
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which can yield a fairly complex conclusion. |
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Here is a simple example (which is proved by @{text"blast"}): |
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*}; |
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lemma simple: "\<forall>y. A y \<longrightarrow> B y \<longrightarrow> B y & A y"; |
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(*<*)by blast;(*>*) |
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text{*\noindent |
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You can get the desired lemma by explicit |
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application of modus ponens and @{thm[source]spec}: |
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*}; |
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lemmas myrule = simple[THEN spec, THEN mp, THEN mp]; |
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text{*\noindent |
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or the wholesale stripping of @{text"\<forall>"} and |
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@{text"\<longrightarrow>"} in the conclusion via @{text"rule_format"} |
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*}; |
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lemmas myrule = simple[rule_format]; |
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text{*\noindent |
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yielding @{thm"myrule"[no_vars]}. |
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You can go one step further and include these derivations already in the |
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statement of your original lemma, thus avoiding the intermediate step: |
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*}; |
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lemma myrule[rule_format]: "\<forall>y. A y \<longrightarrow> B y \<longrightarrow> B y & A y"; |
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(*<*) |
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by blast; |
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(*>*) |
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text{* |
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\bigskip |
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A second reason why your proposition may not be amenable to induction is that |
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you want to induct on a whole term, rather than an individual variable. In |
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general, when inducting on some term $t$ you must rephrase the conclusion as |
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\[ \forall y@1 \dots y@n.~ x = t \longrightarrow C \] where $y@1 \dots y@n$ |
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are the free variables in $t$ and $x$ is new, and perform induction on $x$ |
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afterwards. An example appears below. |
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*}; |
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subsection{*Beyond structural and recursion induction*}; |
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text{* |
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So far, inductive proofs where by structural induction for |
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primitive recursive functions and recursion induction for total recursive |
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functions. But sometimes structural induction is awkward and there is no |
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recursive function in sight either that could furnish a more appropriate |
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induction schema. In such cases some existing standard induction schema can |
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be helpful. We show how to apply such induction schemas by an example. |
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Structural induction on @{typ"nat"} is |
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usually known as ``mathematical induction''. There is also ``complete |
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induction'', where you must prove $P(n)$ under the assumption that $P(m)$ |
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holds for all $m<n$. In Isabelle, this is the theorem @{thm[source]nat_less_induct}: |
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@{thm[display]"nat_less_induct"[no_vars]} |
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Here is an example of its application. |
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*}; |
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consts f :: "nat => nat"; |
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axioms f_ax: "f(f(n)) < f(Suc(n))"; |
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text{*\noindent |
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From the above axiom\footnote{In general, the use of axioms is strongly |
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discouraged, because of the danger of inconsistencies. The above axiom does |
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not introduce an inconsistency because, for example, the identity function |
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satisfies it.} |
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for @{term"f"} it follows that @{prop"n <= f n"}, which can |
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be proved by induction on @{term"f n"}. Following the recipy outlined |
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above, we have to phrase the proposition as follows to allow induction: |
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*}; |
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lemma f_incr_lem: "\<forall>i. k = f i \<longrightarrow> i \<le> f i"; |
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txt{*\noindent |
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To perform induction on @{term"k"} using @{thm[source]nat_less_induct}, we use the same |
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general induction method as for recursion induction (see |
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\S\ref{sec:recdef-induction}): |
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*}; |
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apply(induct_tac k rule: nat_less_induct); |
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(*<*) |
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apply(rule allI); |
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apply(case_tac i); |
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apply(simp); |
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(*>*) |
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txt{*\noindent |
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which leaves us with the following proof state: |
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\begin{isabelle} |
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\ 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 |
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\ \ \ \ \ \ \ {\isasymLongrightarrow}\ {\isasymforall}\mbox{i}.\ \mbox{n}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i} |
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\end{isabelle} |
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After stripping the @{text"\<forall>i"}, the proof continues with a case |
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distinction on @{term"i"}. The case @{prop"i = 0"} is trivial and we focus on |
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the other case: |
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\begin{isabelle} |
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\ 1.\ {\isasymAnd}\mbox{n}\ \mbox{i}\ \mbox{nat}.\isanewline |
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\ \ \ \ \ \ \ {\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 |
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\ \ \ \ \ \ \ {\isasymLongrightarrow}\ \mbox{n}\ =\ f\ \mbox{i}\ {\isasymlongrightarrow}\ \mbox{i}\ {\isasymle}\ f\ \mbox{i} |
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\end{isabelle} |
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*}; |
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by(blast intro!: f_ax Suc_leI intro: le_less_trans); |
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text{*\noindent |
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It is not surprising if you find the last step puzzling. |
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The proof goes like this (writing @{term"j"} instead of @{typ"nat"}). |
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Since @{prop"i = Suc j"} it suffices to show |
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@{prop"j < f(Suc j)"} (by @{thm[source]Suc_leI}: @{thm"Suc_leI"[no_vars]}). This is |
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proved as follows. From @{thm[source]f_ax} we have @{prop"f (f j) < f (Suc j)"} |
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(1) which implies @{prop"f j <= f (f j)"} (by the induction hypothesis). |
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Using (1) once more we obtain @{prop"f j < f(Suc j)"} (2) by transitivity |
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(@{thm[source]le_less_trans}: @{thm"le_less_trans"[no_vars]}). |
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Using the induction hypothesis once more we obtain @{prop"j <= f j"} |
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which, together with (2) yields @{prop"j < f (Suc j)"} (again by |
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@{thm[source]le_less_trans}). |
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This last step shows both the power and the danger of automatic proofs: they |
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will usually not tell you how the proof goes, because it can be very hard to |
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translate the internal proof into a human-readable format. Therefore |
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\S\ref{sec:part2?} introduces a language for writing readable yet concise |
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proofs. |
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We can now derive the desired @{prop"i <= f i"} from @{text"f_incr"}: |
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*}; |
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lemmas f_incr = f_incr_lem[rule_format, OF refl]; |
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text{*\noindent |
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The final @{thm[source]refl} gets rid of the premise @{text"?k = f ?i"}. Again, |
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we could have included this derivation in the original statement of the lemma: |
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*}; |
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lemma f_incr[rule_format, OF refl]: "\<forall>i. k = f i \<longrightarrow> i \<le> f i"; |
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(*<*)oops;(*>*) |
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text{* |
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\begin{exercise} |
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From the above axiom and lemma for @{term"f"} show that @{term"f"} is the |
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identity. |
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\end{exercise} |
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In general, @{text"induct_tac"} can be applied with any rule $r$ |
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whose conclusion is of the form ${?}P~?x@1 \dots ?x@n$, in which case the |
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format is |
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\begin{quote} |
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\isacommand{apply}@{text"(induct_tac"} $y@1 \dots y@n$ @{text"rule:"} $r$@{text")"} |
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\end{quote}\index{*induct_tac}% |
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where $y@1, \dots, y@n$ are variables in the first subgoal. |
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In fact, @{text"induct_tac"} even allows the conclusion of |
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$r$ to be an (iterated) conjunction of formulae of the above form, in |
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which case the application is |
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\begin{quote} |
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\isacommand{apply}@{text"(induct_tac"} $y@1 \dots y@n$ @{text"and"} \dots\ @{text"and"} $z@1 \dots z@m$ @{text"rule:"} $r$@{text")"} |
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\end{quote} |
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*}; |
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subsection{*Derivation of new induction schemas*}; |
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text{*\label{sec:derive-ind} |
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Induction schemas are ordinary theorems and you can derive new ones |
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whenever you wish. This section shows you how to, using the example |
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of @{thm[source]nat_less_induct}. Assume we only have structural induction |
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available for @{typ"nat"} and want to derive complete induction. This |
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requires us to generalize the statement first: |
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*}; |
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lemma induct_lem: "(\<And>n::nat. \<forall>m<n. P m \<Longrightarrow> P n) \<Longrightarrow> \<forall>m<n. P m"; |
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apply(induct_tac n); |
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txt{*\noindent |
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The base case is trivially true. For the induction step (@{prop"m < |
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Suc n"}) we distinguish two cases: case @{prop"m < n"} is true by induction |
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hypothesis and case @{prop"m = n"} follows from the assumption, again using |
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the induction hypothesis: |
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*}; |
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apply(blast); |
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by(blast elim:less_SucE) |
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text{*\noindent |
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The elimination rule @{thm[source]less_SucE} expresses the case distinction: |
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@{thm[display]"less_SucE"[no_vars]} |
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Now it is straightforward to derive the original version of |
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@{thm[source]nat_less_induct} by manipulting the conclusion of the above lemma: |
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instantiate @{term"n"} by @{term"Suc n"} and @{term"m"} by @{term"n"} and |
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remove the trivial condition @{prop"n < Sc n"}. Fortunately, this |
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happens automatically when we add the lemma as a new premise to the |
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desired goal: |
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*}; |
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theorem nat_less_induct: "(\<And>n::nat. \<forall>m<n. P m \<Longrightarrow> P n) \<Longrightarrow> P n"; |
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by(insert induct_lem, blast); |
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text{* |
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Finally we should mention that HOL already provides the mother of all |
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inductions, \textbf{wellfounded |
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induction}\indexbold{induction!wellfounded}\index{wellfounded |
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induction|see{induction, wellfounded}} (@{thm[source]wf_induct}): |
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@{thm[display]wf_induct[no_vars]} |
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where @{term"wf r"} means that the relation @{term r} is wellfounded |
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(see \S\ref{sec:wellfounded}). |
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For example, theorem @{thm[source]nat_less_induct} can be viewed (and |
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derived) as a special case of @{thm[source]wf_induct} where |
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@{term r} is @{text"<"} on @{typ nat}. The details can be found in the HOL library. |
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For a mathematical account of wellfounded induction see, for example, \cite{Baader-Nipkow}. |
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*}; |
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(*<*) |
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end |
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(*>*) |