--- a/doc-src/TutorialI/Misc/document/AdvancedInd.tex Thu Jul 26 16:08:16 2012 +0200
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-%
-\begin{isabellebody}%
-\def\isabellecontext{AdvancedInd}%
-%
-\isadelimtheory
-%
-\endisadelimtheory
-%
-\isatagtheory
-%
-\endisatagtheory
-{\isafoldtheory}%
-%
-\isadelimtheory
-%
-\endisadelimtheory
-%
-\begin{isamarkuptext}%
-\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}).%
-\end{isamarkuptext}%
-\isamarkuptrue%
-%
-\isamarkupsubsection{Massaging the Proposition%
-}
-\isamarkuptrue%
-%
-\begin{isamarkuptext}%
-\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 \isa{hd} and \isa{last} return the first and last element of a
-non-empty list, this lemma looks easy to prove:%
-\end{isamarkuptext}%
-\isamarkuptrue%
-\isacommand{lemma}\isamarkupfalse%
-\ {\isaliteral{22}{\isachardoublequoteopen}}xs\ {\isaliteral{5C3C6E6F7465713E}{\isasymnoteq}}\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ hd{\isaliteral{28}{\isacharparenleft}}rev\ xs{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{3D}{\isacharequal}}\ last\ xs{\isaliteral{22}{\isachardoublequoteclose}}\isanewline
-%
-\isadelimproof
-%
-\endisadelimproof
-%
-\isatagproof
-\isacommand{apply}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}induct{\isaliteral{5F}{\isacharunderscore}}tac\ xs{\isaliteral{29}{\isacharparenright}}%
-\begin{isamarkuptxt}%
-\noindent
-But induction produces the warning
-\begin{quote}\tt
-Induction variable occurs also among premises!
-\end{quote}
-and leads to the base case
-\begin{isabelle}%
-\ {\isadigit{1}}{\isaliteral{2E}{\isachardot}}\ xs\ {\isaliteral{5C3C6E6F7465713E}{\isasymnoteq}}\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ hd\ {\isaliteral{28}{\isacharparenleft}}rev\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{3D}{\isacharequal}}\ last\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}%
-\end{isabelle}
-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 \isa{hd} and
-\isa{last} return when applied to \isa{{\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}}.
-
-We should not have ignored the warning. Because the induction
-formula is only the conclusion, induction does not affect the occurrence of \isa{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 \isa{{\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}}.}
-\end{quote}
-Thus we should state the lemma as an ordinary
-implication~(\isa{{\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}}), letting
-\attrdx{rule_format} (\S\ref{sec:forward}) convert the
-result to the usual \isa{{\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}} form:%
-\end{isamarkuptxt}%
-\isamarkuptrue%
-%
-\endisatagproof
-{\isafoldproof}%
-%
-\isadelimproof
-%
-\endisadelimproof
-\isacommand{lemma}\isamarkupfalse%
-\ hd{\isaliteral{5F}{\isacharunderscore}}rev\ {\isaliteral{5B}{\isacharbrackleft}}rule{\isaliteral{5F}{\isacharunderscore}}format{\isaliteral{5D}{\isacharbrackright}}{\isaliteral{3A}{\isacharcolon}}\ {\isaliteral{22}{\isachardoublequoteopen}}xs\ {\isaliteral{5C3C6E6F7465713E}{\isasymnoteq}}\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ hd{\isaliteral{28}{\isacharparenleft}}rev\ xs{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{3D}{\isacharequal}}\ last\ xs{\isaliteral{22}{\isachardoublequoteclose}}%
-\isadelimproof
-%
-\endisadelimproof
-%
-\isatagproof
-%
-\begin{isamarkuptxt}%
-\noindent
-This time, induction leaves us with a trivial base case:
-\begin{isabelle}%
-\ {\isadigit{1}}{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}\ {\isaliteral{5C3C6E6F7465713E}{\isasymnoteq}}\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ hd\ {\isaliteral{28}{\isacharparenleft}}rev\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{3D}{\isacharequal}}\ last\ {\isaliteral{5B}{\isacharbrackleft}}{\isaliteral{5D}{\isacharbrackright}}%
-\end{isabelle}
-And \isa{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, \isa{rule{\isaliteral{5F}{\isacharunderscore}}format}
-can remove any number of occurrences of \isa{{\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}} and
-\isa{{\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}}.
-
-\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 \isa{{\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}}, again using \isa{{\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}} 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.%
-\end{isamarkuptxt}%
-\isamarkuptrue%
-%
-\endisatagproof
-{\isafoldproof}%
-%
-\isadelimproof
-%
-\endisadelimproof
-%
-\isamarkupsubsection{Beyond Structural and Recursion Induction%
-}
-\isamarkuptrue%
-%
-\begin{isamarkuptext}%
-\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 \isa{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}:
-\begin{isabelle}%
-\ \ \ \ \ {\isaliteral{28}{\isacharparenleft}}{\isaliteral{5C3C416E643E}{\isasymAnd}}n{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}m{\isaliteral{3C}{\isacharless}}n{\isaliteral{2E}{\isachardot}}\ P\ m\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ n{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ n%
-\end{isabelle}
-As an application, we prove a property of the following
-function:%
-\end{isamarkuptext}%
-\isamarkuptrue%
-\isacommand{consts}\isamarkupfalse%
-\ f\ {\isaliteral{3A}{\isacharcolon}}{\isaliteral{3A}{\isacharcolon}}\ {\isaliteral{22}{\isachardoublequoteopen}}nat\ {\isaliteral{5C3C52696768746172726F773E}{\isasymRightarrow}}\ nat{\isaliteral{22}{\isachardoublequoteclose}}\isanewline
-\isacommand{axioms}\isamarkupfalse%
-\ f{\isaliteral{5F}{\isacharunderscore}}ax{\isaliteral{3A}{\isacharcolon}}\ {\isaliteral{22}{\isachardoublequoteopen}}f{\isaliteral{28}{\isacharparenleft}}f{\isaliteral{28}{\isacharparenleft}}n{\isaliteral{29}{\isacharparenright}}{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{3C}{\isacharless}}\ f{\isaliteral{28}{\isacharparenleft}}Suc{\isaliteral{28}{\isacharparenleft}}n{\isaliteral{29}{\isacharparenright}}{\isaliteral{29}{\isacharparenright}}{\isaliteral{22}{\isachardoublequoteclose}}%
-\begin{isamarkuptext}%
-\begin{warn}
-We discourage the use of axioms because of the danger of
-inconsistencies. Axiom \isa{f{\isaliteral{5F}{\isacharunderscore}}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 \isa{f} implies \isa{n\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ n}, which can
-be proved by induction on \mbox{\isa{f\ n}}. Following the recipe outlined
-above, we have to phrase the proposition as follows to allow induction:%
-\end{isamarkuptext}%
-\isamarkuptrue%
-\isacommand{lemma}\isamarkupfalse%
-\ f{\isaliteral{5F}{\isacharunderscore}}incr{\isaliteral{5F}{\isacharunderscore}}lem{\isaliteral{3A}{\isacharcolon}}\ {\isaliteral{22}{\isachardoublequoteopen}}{\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}i{\isaliteral{2E}{\isachardot}}\ k\ {\isaliteral{3D}{\isacharequal}}\ f\ i\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ i\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ i{\isaliteral{22}{\isachardoublequoteclose}}%
-\isadelimproof
-%
-\endisadelimproof
-%
-\isatagproof
-%
-\begin{isamarkuptxt}%
-\noindent
-To perform induction on \isa{k} using \isa{nat{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}induct}, we use
-the same general induction method as for recursion induction (see
-\S\ref{sec:fun-induction}):%
-\end{isamarkuptxt}%
-\isamarkuptrue%
-\isacommand{apply}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}induct{\isaliteral{5F}{\isacharunderscore}}tac\ k\ rule{\isaliteral{3A}{\isacharcolon}}\ nat{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}induct{\isaliteral{29}{\isacharparenright}}%
-\begin{isamarkuptxt}%
-\noindent
-We get the following proof state:
-\begin{isabelle}%
-\ {\isadigit{1}}{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C416E643E}{\isasymAnd}}n{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}m{\isaliteral{3C}{\isacharless}}n{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}i{\isaliteral{2E}{\isachardot}}\ m\ {\isaliteral{3D}{\isacharequal}}\ f\ i\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ i\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ i\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}i{\isaliteral{2E}{\isachardot}}\ n\ {\isaliteral{3D}{\isacharequal}}\ f\ i\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ i\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ i%
-\end{isabelle}
-After stripping the \isa{{\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}i}, the proof continues with a case
-distinction on \isa{i}. The case \isa{i\ {\isaliteral{3D}{\isacharequal}}\ {\isadigit{0}}} is trivial and we focus on
-the other case:%
-\end{isamarkuptxt}%
-\isamarkuptrue%
-\isacommand{apply}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}rule\ allI{\isaliteral{29}{\isacharparenright}}\isanewline
-\isacommand{apply}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}case{\isaliteral{5F}{\isacharunderscore}}tac\ i{\isaliteral{29}{\isacharparenright}}\isanewline
-\ \isacommand{apply}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}simp{\isaliteral{29}{\isacharparenright}}%
-\begin{isamarkuptxt}%
-\begin{isabelle}%
-\ {\isadigit{1}}{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C416E643E}{\isasymAnd}}n\ i\ nat{\isaliteral{2E}{\isachardot}}\isanewline
-\isaindent{\ {\isadigit{1}}{\isaliteral{2E}{\isachardot}}\ \ \ \ }{\isaliteral{5C3C6C6272616B6B3E}{\isasymlbrakk}}{\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}m{\isaliteral{3C}{\isacharless}}n{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}i{\isaliteral{2E}{\isachardot}}\ m\ {\isaliteral{3D}{\isacharequal}}\ f\ i\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ i\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ i{\isaliteral{3B}{\isacharsemicolon}}\ i\ {\isaliteral{3D}{\isacharequal}}\ Suc\ nat{\isaliteral{5C3C726272616B6B3E}{\isasymrbrakk}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ n\ {\isaliteral{3D}{\isacharequal}}\ f\ i\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ i\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ i%
-\end{isabelle}%
-\end{isamarkuptxt}%
-\isamarkuptrue%
-\isacommand{by}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}blast\ intro{\isaliteral{21}{\isacharbang}}{\isaliteral{3A}{\isacharcolon}}\ f{\isaliteral{5F}{\isacharunderscore}}ax\ Suc{\isaliteral{5F}{\isacharunderscore}}leI\ intro{\isaliteral{3A}{\isacharcolon}}\ le{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}trans{\isaliteral{29}{\isacharparenright}}%
-\endisatagproof
-{\isafoldproof}%
-%
-\isadelimproof
-%
-\endisadelimproof
-%
-\begin{isamarkuptext}%
-\noindent
-If you find the last step puzzling, here are the two lemmas it employs:
-\begin{isabelle}
-\isa{m\ {\isaliteral{3C}{\isacharless}}\ n\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ Suc\ m\ {\isaliteral{5C3C6C653E}{\isasymle}}\ n}
-\rulename{Suc_leI}\isanewline
-\isa{{\isaliteral{5C3C6C6272616B6B3E}{\isasymlbrakk}}x\ {\isaliteral{5C3C6C653E}{\isasymle}}\ y{\isaliteral{3B}{\isacharsemicolon}}\ y\ {\isaliteral{3C}{\isacharless}}\ z{\isaliteral{5C3C726272616B6B3E}{\isasymrbrakk}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ x\ {\isaliteral{3C}{\isacharless}}\ z}
-\rulename{le_less_trans}
-\end{isabelle}
-%
-The proof goes like this (writing \isa{j} instead of \isa{nat}).
-Since \isa{i\ {\isaliteral{3D}{\isacharequal}}\ Suc\ j} it suffices to show
-\hbox{\isa{j\ {\isaliteral{3C}{\isacharless}}\ f\ {\isaliteral{28}{\isacharparenleft}}Suc\ j{\isaliteral{29}{\isacharparenright}}}},
-by \isa{Suc{\isaliteral{5F}{\isacharunderscore}}leI}\@. This is
-proved as follows. From \isa{f{\isaliteral{5F}{\isacharunderscore}}ax} we have \isa{f\ {\isaliteral{28}{\isacharparenleft}}f\ j{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{3C}{\isacharless}}\ f\ {\isaliteral{28}{\isacharparenleft}}Suc\ j{\isaliteral{29}{\isacharparenright}}}
-(1) which implies \isa{f\ j\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ {\isaliteral{28}{\isacharparenleft}}f\ j{\isaliteral{29}{\isacharparenright}}} by the induction hypothesis.
-Using (1) once more we obtain \isa{f\ j\ {\isaliteral{3C}{\isacharless}}\ f\ {\isaliteral{28}{\isacharparenleft}}Suc\ j{\isaliteral{29}{\isacharparenright}}} (2) by the transitivity
-rule \isa{le{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}trans}.
-Using the induction hypothesis once more we obtain \isa{j\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ j}
-which, together with (2) yields \isa{j\ {\isaliteral{3C}{\isacharless}}\ f\ {\isaliteral{28}{\isacharparenleft}}Suc\ j{\isaliteral{29}{\isacharparenright}}} (again by
-\isa{le{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}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, \isa{i\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ i}, follows from \isa{f{\isaliteral{5F}{\isacharunderscore}}incr{\isaliteral{5F}{\isacharunderscore}}lem}:%
-\end{isamarkuptext}%
-\isamarkuptrue%
-\isacommand{lemmas}\isamarkupfalse%
-\ f{\isaliteral{5F}{\isacharunderscore}}incr\ {\isaliteral{3D}{\isacharequal}}\ f{\isaliteral{5F}{\isacharunderscore}}incr{\isaliteral{5F}{\isacharunderscore}}lem{\isaliteral{5B}{\isacharbrackleft}}rule{\isaliteral{5F}{\isacharunderscore}}format{\isaliteral{2C}{\isacharcomma}}\ OF\ refl{\isaliteral{5D}{\isacharbrackright}}%
-\begin{isamarkuptext}%
-\noindent
-The final \isa{refl} gets rid of the premise \isa{{\isaliteral{3F}{\isacharquery}}k\ {\isaliteral{3D}{\isacharequal}}\ f\ {\isaliteral{3F}{\isacharquery}}i}.
-We could have included this derivation in the original statement of the lemma:%
-\end{isamarkuptext}%
-\isamarkuptrue%
-\isacommand{lemma}\isamarkupfalse%
-\ f{\isaliteral{5F}{\isacharunderscore}}incr{\isaliteral{5B}{\isacharbrackleft}}rule{\isaliteral{5F}{\isacharunderscore}}format{\isaliteral{2C}{\isacharcomma}}\ OF\ refl{\isaliteral{5D}{\isacharbrackright}}{\isaliteral{3A}{\isacharcolon}}\ {\isaliteral{22}{\isachardoublequoteopen}}{\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}i{\isaliteral{2E}{\isachardot}}\ k\ {\isaliteral{3D}{\isacharequal}}\ f\ i\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ i\ {\isaliteral{5C3C6C653E}{\isasymle}}\ f\ i{\isaliteral{22}{\isachardoublequoteclose}}%
-\isadelimproof
-%
-\endisadelimproof
-%
-\isatagproof
-%
-\endisatagproof
-{\isafoldproof}%
-%
-\isadelimproof
-%
-\endisadelimproof
-%
-\begin{isamarkuptext}%
-\begin{exercise}
-From the axiom and lemma for \isa{f}, show that \isa{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}\isa{{\isaliteral{28}{\isacharparenleft}}induct{\isaliteral{5F}{\isacharunderscore}}tac} $y@1 \dots y@n$ \isa{rule{\isaliteral{3A}{\isacharcolon}}} $r$\isa{{\isaliteral{29}{\isacharparenright}}}
-\end{quote}
-where $y@1, \dots, y@n$ are variables in the conclusion of the first subgoal.
-
-A further useful induction rule is \isa{length{\isaliteral{5F}{\isacharunderscore}}induct},
-induction on the length of a list\indexbold{*length_induct}
-\begin{isabelle}%
-\ \ \ \ \ {\isaliteral{28}{\isacharparenleft}}{\isaliteral{5C3C416E643E}{\isasymAnd}}xs{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}ys{\isaliteral{2E}{\isachardot}}\ length\ ys\ {\isaliteral{3C}{\isacharless}}\ length\ xs\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ P\ ys\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ xs{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ xs%
-\end{isabelle}
-which is a special case of \isa{measure{\isaliteral{5F}{\isacharunderscore}}induct}
-\begin{isabelle}%
-\ \ \ \ \ {\isaliteral{28}{\isacharparenleft}}{\isaliteral{5C3C416E643E}{\isasymAnd}}x{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}y{\isaliteral{2E}{\isachardot}}\ f\ y\ {\isaliteral{3C}{\isacharless}}\ f\ x\ {\isaliteral{5C3C6C6F6E6772696768746172726F773E}{\isasymlongrightarrow}}\ P\ y\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ x{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ a%
-\end{isabelle}
-where \isa{f} may be any function into type \isa{nat}.%
-\end{isamarkuptext}%
-\isamarkuptrue%
-%
-\isamarkupsubsection{Derivation of New Induction Schemas%
-}
-\isamarkuptrue%
-%
-\begin{isamarkuptext}%
-\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 \isa{nat{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}induct}. Assume we only have structural induction
-available for \isa{nat} and want to derive complete induction. We
-must generalize the statement as shown:%
-\end{isamarkuptext}%
-\isamarkuptrue%
-\isacommand{lemma}\isamarkupfalse%
-\ induct{\isaliteral{5F}{\isacharunderscore}}lem{\isaliteral{3A}{\isacharcolon}}\ {\isaliteral{22}{\isachardoublequoteopen}}{\isaliteral{28}{\isacharparenleft}}{\isaliteral{5C3C416E643E}{\isasymAnd}}n{\isaliteral{3A}{\isacharcolon}}{\isaliteral{3A}{\isacharcolon}}nat{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}m{\isaliteral{3C}{\isacharless}}n{\isaliteral{2E}{\isachardot}}\ P\ m\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ n{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}m{\isaliteral{3C}{\isacharless}}n{\isaliteral{2E}{\isachardot}}\ P\ m{\isaliteral{22}{\isachardoublequoteclose}}\isanewline
-%
-\isadelimproof
-%
-\endisadelimproof
-%
-\isatagproof
-\isacommand{apply}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}induct{\isaliteral{5F}{\isacharunderscore}}tac\ n{\isaliteral{29}{\isacharparenright}}%
-\begin{isamarkuptxt}%
-\noindent
-The base case is vacuously true. For the induction step (\isa{m\ {\isaliteral{3C}{\isacharless}}\ Suc\ n}) we distinguish two cases: case \isa{m\ {\isaliteral{3C}{\isacharless}}\ n} is true by induction
-hypothesis and case \isa{m\ {\isaliteral{3D}{\isacharequal}}\ n} follows from the assumption, again using
-the induction hypothesis:%
-\end{isamarkuptxt}%
-\isamarkuptrue%
-\ \isacommand{apply}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}blast{\isaliteral{29}{\isacharparenright}}\isanewline
-\isacommand{by}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}blast\ elim{\isaliteral{3A}{\isacharcolon}}\ less{\isaliteral{5F}{\isacharunderscore}}SucE{\isaliteral{29}{\isacharparenright}}%
-\endisatagproof
-{\isafoldproof}%
-%
-\isadelimproof
-%
-\endisadelimproof
-%
-\begin{isamarkuptext}%
-\noindent
-The elimination rule \isa{less{\isaliteral{5F}{\isacharunderscore}}SucE} expresses the case distinction:
-\begin{isabelle}%
-\ \ \ \ \ {\isaliteral{5C3C6C6272616B6B3E}{\isasymlbrakk}}m\ {\isaliteral{3C}{\isacharless}}\ Suc\ n{\isaliteral{3B}{\isacharsemicolon}}\ m\ {\isaliteral{3C}{\isacharless}}\ n\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P{\isaliteral{3B}{\isacharsemicolon}}\ m\ {\isaliteral{3D}{\isacharequal}}\ n\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P{\isaliteral{5C3C726272616B6B3E}{\isasymrbrakk}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P%
-\end{isabelle}
-
-Now it is straightforward to derive the original version of
-\isa{nat{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}induct} by manipulating the conclusion of the above
-lemma: instantiate \isa{n} by \isa{Suc\ n} and \isa{m} by \isa{n}
-and remove the trivial condition \isa{n\ {\isaliteral{3C}{\isacharless}}\ Suc\ n}. Fortunately, this
-happens automatically when we add the lemma as a new premise to the
-desired goal:%
-\end{isamarkuptext}%
-\isamarkuptrue%
-\isacommand{theorem}\isamarkupfalse%
-\ nat{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}induct{\isaliteral{3A}{\isacharcolon}}\ {\isaliteral{22}{\isachardoublequoteopen}}{\isaliteral{28}{\isacharparenleft}}{\isaliteral{5C3C416E643E}{\isasymAnd}}n{\isaliteral{3A}{\isacharcolon}}{\isaliteral{3A}{\isacharcolon}}nat{\isaliteral{2E}{\isachardot}}\ {\isaliteral{5C3C666F72616C6C3E}{\isasymforall}}m{\isaliteral{3C}{\isacharless}}n{\isaliteral{2E}{\isachardot}}\ P\ m\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ n{\isaliteral{29}{\isacharparenright}}\ {\isaliteral{5C3C4C6F6E6772696768746172726F773E}{\isasymLongrightarrow}}\ P\ n{\isaliteral{22}{\isachardoublequoteclose}}\isanewline
-%
-\isadelimproof
-%
-\endisadelimproof
-%
-\isatagproof
-\isacommand{by}\isamarkupfalse%
-{\isaliteral{28}{\isacharparenleft}}insert\ induct{\isaliteral{5F}{\isacharunderscore}}lem{\isaliteral{2C}{\isacharcomma}}\ blast{\isaliteral{29}{\isacharparenright}}%
-\endisatagproof
-{\isafoldproof}%
-%
-\isadelimproof
-%
-\endisadelimproof
-%
-\begin{isamarkuptext}%
-HOL already provides the mother of
-all inductions, well-founded induction (see \S\ref{sec:Well-founded}). For
-example theorem \isa{nat{\isaliteral{5F}{\isacharunderscore}}less{\isaliteral{5F}{\isacharunderscore}}induct} is
-a special case of \isa{wf{\isaliteral{5F}{\isacharunderscore}}induct} where \isa{r} is \isa{{\isaliteral{3C}{\isacharless}}} on
-\isa{nat}. The details can be found in theory \isa{Wellfounded_Recursion}.%
-\end{isamarkuptext}%
-\isamarkuptrue%
-%
-\isadelimtheory
-%
-\endisadelimtheory
-%
-\isatagtheory
-%
-\endisatagtheory
-{\isafoldtheory}%
-%
-\isadelimtheory
-%
-\endisadelimtheory
-\end{isabellebody}%
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