theory natsum imports Main begin
text‹\noindent
In particular, there are @{text"case"}-expressions, for example
@{term[display]"case n of 0 => 0 | Suc m => m"}
primitive recursion, for example
›
primrec sum :: "nat ⇒ nat" where
"sum 0 = 0" |
"sum (Suc n) = Suc n + sum n"
text‹\noindent
and induction, for example
›
lemma "sum n + sum n = n*(Suc n)"
apply(induct_tac n)
apply(auto)
done
text‹\newcommand{\mystar}{*%
}
\index{arithmetic operations!for \protect\isa{nat}}%
The arithmetic operations \isadxboldpos{+}{$HOL2arithfun},
\isadxboldpos{-}{$HOL2arithfun}, \isadxboldpos{\mystar}{$HOL2arithfun},
\sdx{div}, \sdx{mod}, \cdx{min} and
\cdx{max} are predefined, as are the relations
\isadxboldpos{\isasymle}{$HOL2arithrel} and
\isadxboldpos{<}{$HOL2arithrel}. As usual, @{prop"m-n = (0::nat)"} if
@{prop"m<n"}. There is even a least number operation
\sdx{LEAST}\@. For example, @{prop"(LEAST n. 0 < n) = Suc 0"}.
\begin{warn}\index{overloading}
The constants \cdx{0} and \cdx{1} and the operations
\isadxboldpos{+}{$HOL2arithfun}, \isadxboldpos{-}{$HOL2arithfun},
\isadxboldpos{\mystar}{$HOL2arithfun}, \cdx{min},
\cdx{max}, \isadxboldpos{\isasymle}{$HOL2arithrel} and
\isadxboldpos{<}{$HOL2arithrel} are overloaded: they are available
not just for natural numbers but for other types as well.
For example, given the goal @{text"x + 0 = x"}, there is nothing to indicate
that you are talking about natural numbers. Hence Isabelle can only infer
that @{term x} is of some arbitrary type where @{text 0} and @{text"+"} are
declared. As a consequence, you will be unable to prove the
goal. To alert you to such pitfalls, Isabelle flags numerals without a
fixed type in its output: @{prop"x+0 = x"}. (In the absence of a numeral,
it may take you some time to realize what has happened if \pgmenu{Show
Types} is not set). In this particular example, you need to include
an explicit type constraint, for example @{text"x+0 = (x::nat)"}. If there
is enough contextual information this may not be necessary: @{prop"Suc x =
x"} automatically implies @{text"x::nat"} because @{term Suc} is not
overloaded.
For details on overloading see \S\ref{sec:overloading}.
Table~\ref{tab:overloading} in the appendix shows the most important
overloaded operations.
\end{warn}
\begin{warn}
The symbols \isadxboldpos{>}{$HOL2arithrel} and
\isadxboldpos{\isasymge}{$HOL2arithrel} are merely syntax: @{text"x > y"}
stands for @{prop"y < x"} and similary for @{text"≥"} and
@{text"≤"}.
\end{warn}
\begin{warn}
Constant @{text"1::nat"} is defined to equal @{term"Suc 0"}. This definition
(see \S\ref{sec:ConstDefinitions}) is unfolded automatically by some
tactics (like @{text auto}, @{text simp} and @{text arith}) but not by
others (especially the single step tactics in Chapter~\ref{chap:rules}).
If you need the full set of numerals, see~\S\ref{sec:numerals}.
\emph{Novices are advised to stick to @{term"0::nat"} and @{term Suc}.}
\end{warn}
Both @{text auto} and @{text simp}
(a method introduced below, \S\ref{sec:Simplification}) prove
simple arithmetic goals automatically:
›
lemma "⟦ ¬ m < n; m < n + (1::nat) ⟧ ⟹ m = n"
by(auto)
text‹\noindent
For efficiency's sake, this built-in prover ignores quantified formulae,
many logical connectives, and all arithmetic operations apart from addition.
In consequence, @{text auto} and @{text simp} cannot prove this slightly more complex goal:
›
lemma "m ≠ (n::nat) ⟹ m < n ∨ n < m"
by(arith)
text‹\noindent The method \methdx{arith} is more general. It attempts to
prove the first subgoal provided it is a \textbf{linear arithmetic} formula.
Such formulas may involve the usual logical connectives (@{text"¬"},
@{text"∧"}, @{text"∨"}, @{text"⟶"}, @{text"="},
@{text"∀"}, @{text"∃"}), the relations @{text"="},
@{text"≤"} and @{text"<"}, and the operations @{text"+"}, @{text"-"},
@{term min} and @{term max}. For example,›
lemma "min i (max j (k*k)) = max (min (k*k) i) (min i (j::nat))"
apply(arith)
done
text‹\noindent
succeeds because @{term"k*k"} can be treated as atomic. In contrast,
›
lemma "n*n = n+1 ⟹ n=0"
oops
text‹\noindent
is not proved by @{text arith} because the proof relies
on properties of multiplication. Only multiplication by numerals (which is
the same as iterated addition) is taken into account.
\begin{warn} The running time of @{text arith} is exponential in the number
of occurrences of \ttindexboldpos{-}{$HOL2arithfun}, \cdx{min} and
\cdx{max} because they are first eliminated by case distinctions.
If @{text k} is a numeral, \sdx{div}~@{text k}, \sdx{mod}~@{text k} and
@{text k}~\sdx{dvd} are also supported, where the former two are eliminated
by case distinctions, again blowing up the running time.
If the formula involves quantifiers, @{text arith} may take
super-exponential time and space.
\end{warn}
›
end