doc-src/TutorialI/Misc/natsum.thy
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(*<*)
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theory natsum = Main:;
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(*>*)
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text{*\noindent
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In particular, there are @{text"case"}-expressions, for example
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@{term[display]"case n of 0 => 0 | Suc m => m"}
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primitive recursion, for example
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*}
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consts sum :: "nat \<Rightarrow> nat";
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primrec "sum 0 = 0"
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        "sum (Suc n) = Suc n + sum n";
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text{*\noindent
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and induction, for example
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*}
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lemma "sum n + sum n = n*(Suc n)";
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apply(induct_tac n);
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apply(auto);
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done
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text{*\newcommand{\mystar}{*%
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}
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\index{arithmetic operations!for \protect\isa{nat}}%
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The usual arithmetic operations \ttindexboldpos{+}{$HOL2arithfun},
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\ttindexboldpos{-}{$HOL2arithfun}, \ttindexboldpos{\mystar}{$HOL2arithfun},
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\sdx{div}, \sdx{mod}, \cdx{min} and
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\cdx{max} are predefined, as are the relations
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\indexboldpos{\isasymle}{$HOL2arithrel} and
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\ttindexboldpos{<}{$HOL2arithrel}. As usual, @{prop"m-n = 0"} if
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@{prop"m<n"}. There is even a least number operation
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\sdx{LEAST}\@.  For example, @{prop"(LEAST n. 1 < n) = 2"}. 
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\REMARK{Isabelle CAN prove it automatically, using \isa{auto intro: Least_equality}.
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 The following needs changing with our new system of numbers.}
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Note that @{term 1}
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and @{term 2} are available as abbreviations for the corresponding
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@{term Suc}-expressions. If you need the full set of numerals,
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see~\S\ref{sec:numerals}.
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\begin{warn}\index{overloading}
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  The constant \cdx{0} and the operations
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  \ttindexboldpos{+}{$HOL2arithfun}, \ttindexboldpos{-}{$HOL2arithfun},
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  \ttindexboldpos{\mystar}{$HOL2arithfun}, \cdx{min},
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  \cdx{max}, \indexboldpos{\isasymle}{$HOL2arithrel} and
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  \ttindexboldpos{<}{$HOL2arithrel} are overloaded, i.e.\ they are available
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  not just for natural numbers but at other types as well.
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  For example, given the goal @{prop"x+0 = x"},
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  there is nothing to indicate that you are talking about natural numbers.
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  Hence Isabelle can only infer that @{term x} is of some arbitrary type where
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  @{term 0} and @{text"+"} are declared. As a consequence, you will be unable
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  to prove the goal (although it may take you some time to realize what has
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  happened if @{text show_types} is not set).  In this particular example,
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  you need to include an explicit type constraint, for example
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  @{text"x+0 = (x::nat)"}. If there is enough contextual information this
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  may not be necessary: @{prop"Suc x = x"} automatically implies
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  @{text"x::nat"} because @{term Suc} is not overloaded.
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  For details see \S\ref{sec:numbers} and \S\ref{sec:overloading};
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  Table~\ref{tab:overloading} in the appendix shows the most important overloaded
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  operations.
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\end{warn}
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Both @{text auto} and @{text simp}
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(a method introduced below, \S\ref{sec:Simplification}) prove 
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simple arithmetic goals automatically:
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*}
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lemma "\<lbrakk> \<not> m < n; m < n+1 \<rbrakk> \<Longrightarrow> m = n"
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(*<*)by(auto)(*>*)
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text{*\noindent
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For efficiency's sake, this built-in prover ignores quantified formulae,
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logical connectives, and all arithmetic operations apart from addition.
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In consequence, @{text auto} cannot prove this slightly more complex goal:
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*}
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lemma "\<not> m < n \<and> m < n+1 \<Longrightarrow> m = n";
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(*<*)by(arith)(*>*)
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text{*\noindent
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The method \methdx{arith} is more general.  It attempts to prove
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the first subgoal provided it is a quantifier-free \textbf{linear arithmetic}
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formula.  Such formulas may involve the
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usual logical connectives (@{text"\<not>"}, @{text"\<and>"}, @{text"\<or>"},
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@{text"\<longrightarrow>"}), the relations @{text"="}, @{text"\<le>"} and @{text"<"},
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and the operations
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@{text"+"}, @{text"-"}, @{term min} and @{term max}. 
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For example,
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*}
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lemma "min i (max j (k*k)) = max (min (k*k) i) (min i (j::nat))";
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apply(arith)
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(*<*)done(*>*)
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text{*\noindent
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succeeds because @{term"k*k"} can be treated as atomic. In contrast,
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*}
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lemma "n*n = n \<Longrightarrow> n=0 \<or> n=1"
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(*<*)oops(*>*)
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text{*\noindent
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is not proved even by @{text arith} because the proof relies 
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on properties of multiplication.
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\begin{warn}
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  The running time of @{text arith} is exponential in the number of occurrences
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  of \ttindexboldpos{-}{$HOL2arithfun}, \cdx{min} and
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  \cdx{max} because they are first eliminated by case distinctions.
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  Even for linear arithmetic formulae, \isa{arith} is incomplete. If divisibility plays a
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  role, it may fail to prove a valid formula, for example
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  @{prop"m+m \<noteq> n+n+1"}. Fortunately, such examples are rare.
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\end{warn}
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*}
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(*<*)
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end
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(*>*)