(*<*)

 theory AdvancedInd imports Main begin;

 (*>*)

 text{*\noindent

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

 finer points of induction.  We consider two questions: what to do if the

 proposition to be proved is not directly amenable to induction

 (\S\ref{sec:ind-var-in-prems}), and how to utilize (\S\ref{sec:complete-ind})

 and even derive (\S\ref{sec:derive-ind}) new induction schemas. We conclude

 with an extended example of induction (\S\ref{sec:CTL-revisited}).

 *};

 subsection{*Massaging the Proposition*};

 text{*\label{sec:ind-var-in-prems}

 Often we have assumed that the theorem to be proved is already in a form

 that is amenable to induction, but sometimes it isn't.

 Here is an example.

 Since @{term"hd"} and @{term"last"} return the first and last element of a

 non-empty list, this lemma looks easy to prove:

 *};

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

 apply(induct_tac xs)

 txt{*\noindent

 But induction produces the warning

 \begin{quote}\tt

 Induction variable occurs also among premises!

 \end{quote}

 and leads to the base case

 @{subgoals[display,indent=0,goals_limit=1]}

 Simplification reduces the base case to this:

 \begin{isabelle}

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

 \end{isabelle}

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

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

 We should not have ignored the warning. Because the induction

 formula is only the conclusion, induction does not affect the occurrence of @{term xs} in the premises.  

 Thus the case that should have been trivial

 becomes unprovable. Fortunately, the solution is easy:\footnote{A similar

 heuristic applies to rule inductions; see \S\ref{sec:rtc}.}

 \begin{quote}

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

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

 \end{quote}

 Thus we should state the lemma as an ordinary 

 implication~(@{text"\<longrightarrow>"}), letting

 \attrdx{rule_format} (\S\ref{sec:forward}) convert the

 result to the usual @{text"\<Longrightarrow>"} form:

 *};

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

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

 (*<*)

 apply(induct_tac xs);

 (*>*)

 txt{*\noindent

 This time, induction leaves us with a trivial base case:

 @{subgoals[display,indent=0,goals_limit=1]}

 And @{text"auto"} completes the proof.

 If there are multiple premises $A@1$, \dots, $A@n$ containing the

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

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

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

 which can yield a fairly complex conclusion.  However, @{text rule_format} 

 can remove any number of occurrences of @{text"\<forall>"} and

 @{text"\<longrightarrow>"}.

 \index{induction!on a term}%

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

 you want to induct on a complex term, rather than a variable. In

 general, induction on a term~$t$ requires rephrasing the conclusion~$C$

 as

 \begin{equation}\label{eqn:ind-over-term}

 \forall y@1 \dots y@n.~ x = t \longrightarrow C.

 \end{equation}

 where $y@1 \dots y@n$ are the free variables in $t$ and $x$ is a new variable.

 Now you can perform induction on~$x$. An example appears in

 \S\ref{sec:complete-ind} below.

 The very same problem may occur in connection with rule induction. Remember

 that it requires a premise of the form $(x@1,\dots,x@k) \in R$, where $R$ is

 some inductively defined set and the $x@i$ are variables.  If instead we have

 a premise $t \in R$, where $t$ is not just an $n$-tuple of variables, we

 replace it with $(x@1,\dots,x@k) \in R$, and rephrase the conclusion $C$ as

 \[ \forall y@1 \dots y@n.~ (x@1,\dots,x@k) = t \longrightarrow C. \]

 For an example see \S\ref{sec:CTL-revisited} below.

 Of course, all premises that share free variables with $t$ need to be pulled into

 the conclusion as well, under the @{text"\<forall>"}, again using @{text"\<longrightarrow>"} as shown above.

 Readers who are puzzled by the form of statement

 (\ref{eqn:ind-over-term}) above should remember that the

 transformation is only performed to permit induction. Once induction

 has been applied, the statement can be transformed back into something quite

 intuitive. For example, applying wellfounded induction on $x$ (w.r.t.\

 $\prec$) to (\ref{eqn:ind-over-term}) and transforming the result a

 little leads to the goal

 \[ \bigwedge\overline{y}.\ 

    \forall \overline{z}.\ t\,\overline{z} \prec t\,\overline{y}\ \longrightarrow\ C\,\overline{z}

     \ \Longrightarrow\ C\,\overline{y} \]

 where $\overline{y}$ stands for $y@1 \dots y@n$ and the dependence of $t$ and

 $C$ on the free variables of $t$ has been made explicit.

 Unfortunately, this induction schema cannot be expressed as a

 single theorem because it depends on the number of free variables in $t$ ---

 the notation $\overline{y}$ is merely an informal device.

 *}

 (*<*)by auto(*>*)

 subsection{*Beyond Structural and Recursion Induction*};

 text{*\label{sec:complete-ind}

 So far, inductive proofs were by structural induction for

 primitive recursive functions and recursion induction for total recursive

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

 recursive function that could furnish a more appropriate

 induction schema. In such cases a general-purpose induction schema can

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

 Structural induction on @{typ"nat"} is

 usually known as mathematical induction. There is also \textbf{complete}

 \index{induction!complete}%

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

 holds for all $m<n$. In Isabelle, this is the theorem \tdx{nat_less_induct}:

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

 As an application, we prove a property of the following

 function:

 *};

 consts f :: "nat \<Rightarrow> nat";

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

 text{*

 \begin{warn}

 We discourage the use of axioms because of the danger of

 inconsistencies.  Axiom @{text f_ax} does

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

 satisfies it.  Axioms can be useful in exploratory developments, say when 

 you assume some well-known theorems so that you can quickly demonstrate some

 point about methodology.  If your example turns into a substantial proof

 development, you should replace axioms by theorems.

 \end{warn}\noindent

 The axiom for @{term"f"} implies @{prop"n <= f n"}, which can

 be proved by induction on \mbox{@{term"f n"}}. Following the recipe outlined

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

 *};

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

 txt{*\noindent

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

 the same general induction method as for recursion induction (see

 \S\ref{sec:fun-induction}):

 *};

 apply(induct_tac k rule: nat_less_induct);

 txt{*\noindent

 We get the following proof state:

 @{subgoals[display,indent=0,margin=65]}

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

 distinction on @{term"i"}. The case @{prop"i = (0::nat)"} is trivial and we focus on

 the other case:

 *}

 apply(rule allI)

 apply(case_tac i)

  apply(simp)

 txt{*

 @{subgoals[display,indent=0]}

 *}

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

 text{*\noindent

 If you find the last step puzzling, here are the two lemmas it employs:

 \begin{isabelle}

 @{thm Suc_leI[no_vars]}

 \rulename{Suc_leI}\isanewline

 @{thm le_less_trans[no_vars]}

 \rulename{le_less_trans}

 \end{isabelle}

 %

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

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

 \hbox{@{prop"j < f(Suc j)"}},

 by @{thm[source]Suc_leI}\@.  This is

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

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

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

 rule @{thm[source]le_less_trans}.

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

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

 @{thm[source]le_less_trans}).

 This last step shows both the power and the danger of automatic proofs.  They

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

 translate the internal proof into a human-readable format.  Automatic

 proofs are easy to write but hard to read and understand.

 The desired result, @{prop"i <= f i"}, follows from @{thm[source]f_incr_lem}:

 *};

 lemmas f_incr = f_incr_lem[rule_format, OF refl];

 text{*\noindent

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

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

 *};

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

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

 text{*

 \begin{exercise}

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

 identity function.

 \end{exercise}

 Method \methdx{induct_tac} can be applied with any rule $r$

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

 format is

 \begin{quote}

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

 \end{quote}

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

 A further useful induction rule is @{thm[source]length_induct},

 induction on the length of a list\indexbold{*length_induct}

 @{thm[display]length_induct[no_vars]}

 which is a special case of @{thm[source]measure_induct}

 @{thm[display]measure_induct[no_vars]}

 where @{term f} may be any function into type @{typ nat}.

 *}

 subsection{*Derivation of New Induction Schemas*};

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

 \index{induction!deriving new schemas}%

 Induction schemas are ordinary theorems and you can derive new ones

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

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

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

 must generalize the statement as shown:

 *};

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

 apply(induct_tac n);

 txt{*\noindent

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

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

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

 the induction hypothesis:

 *};

  apply(blast);

 by(blast elim: less_SucE)

 text{*\noindent

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

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

 Now it is straightforward to derive the original version of

 @{thm[source]nat_less_induct} by manipulating the conclusion of the above

 lemma: instantiate @{term"n"} by @{term"Suc n"} and @{term"m"} by @{term"n"}

 and remove the trivial condition @{prop"n < Suc n"}. Fortunately, this

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

 desired goal:

 *};

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

 by(insert induct_lem, blast);

 text{*

 HOL already provides the mother of

 all inductions, well-founded induction (see \S\ref{sec:Well-founded}).  For

 example theorem @{thm[source]nat_less_induct} is

 a special case of @{thm[source]wf_induct} where @{term r} is @{text"<"} on

 @{typ nat}. The details can be found in theory \isa{Wellfounded_Recursion}.

 *}

 (*<*)end(*>*)