(* Title: HOL/Isar_Examples/Summation.thy
Author: Markus Wenzel
*)
section \<open>Summing natural numbers\<close>
theory Summation
imports Main
begin
text \<open>Subsequently, we prove some summation laws of natural numbers
(including odds, squares, and cubes). These examples demonstrate how plain
natural deduction (including induction) may be combined with calculational
proof.\<close>
subsection \<open>Summation laws\<close>
text \<open>The sum of natural numbers \<open>0 + \<cdots> + n\<close> equals \<open>n \<times> (n + 1)/2\<close>.
Avoiding formal reasoning about division we prove this equation multiplied
by \<open>2\<close>.\<close>
theorem sum_of_naturals:
"2 * (\<Sum>i::nat=0..n. i) = n * (n + 1)"
(is "?P n" is "?S n = _")
proof (induct n)
show "?P 0" by simp
next
fix n have "?S (n + 1) = ?S n + 2 * (n + 1)"
by simp
also assume "?S n = n * (n + 1)"
also have "\<dots> + 2 * (n + 1) = (n + 1) * (n + 2)"
by simp
finally show "?P (Suc n)"
by simp
qed
text \<open>The above proof is a typical instance of mathematical induction. The
main statement is viewed as some \<open>?P n\<close> that is split by the induction
method into base case \<open>?P 0\<close>, and step case \<open>?P n \<Longrightarrow> ?P (Suc n)\<close> for
arbitrary \<open>n\<close>.
The step case is established by a short calculation in forward manner.
Starting from the left-hand side \<open>?S (n + 1)\<close> of the thesis, the final
result is achieved by transformations involving basic arithmetic reasoning
(using the Simplifier). The main point is where the induction hypothesis
\<open>?S n = n \<times> (n + 1)\<close> is introduced in order to replace a certain subterm.
So the ``transitivity'' rule involved here is actual \<^emph>\<open>substitution\<close>. Also
note how the occurrence of ``\dots'' in the subsequent step documents the
position where the right-hand side of the hypothesis got filled in.
\<^medskip>
A further notable point here is integration of calculations with plain
natural deduction. This works so well in Isar for two reasons.
\<^enum> Facts involved in \isakeyword{also}~/ \isakeyword{finally}
calculational chains may be just anything. There is nothing special
about \isakeyword{have}, so the natural deduction element
\isakeyword{assume} works just as well.
\<^enum> There are two \<^emph>\<open>separate\<close> primitives for building natural deduction
contexts: \isakeyword{fix}~\<open>x\<close> and \isakeyword{assume}~\<open>A\<close>. Thus it is
possible to start reasoning with some new ``arbitrary, but fixed''
elements before bringing in the actual assumption. In contrast, natural
deduction is occasionally formalized with basic context elements of the
form \<open>x:A\<close> instead.
\<^medskip>
We derive further summation laws for odds, squares, and cubes as follows.
The basic technique of induction plus calculation is the same as before.\<close>
theorem sum_of_odds:
"(\<Sum>i::nat=0..<n. 2 * i + 1) = n^Suc (Suc 0)"
(is "?P n" is "?S n = _")
proof (induct n)
show "?P 0" by simp
next
fix n
have "?S (n + 1) = ?S n + 2 * n + 1"
by simp
also assume "?S n = n^Suc (Suc 0)"
also have "\<dots> + 2 * n + 1 = (n + 1)^Suc (Suc 0)"
by simp
finally show "?P (Suc n)"
by simp
qed
text \<open>Subsequently we require some additional tweaking of Isabelle built-in
arithmetic simplifications, such as bringing in distributivity by hand.\<close>
lemmas distrib = add_mult_distrib add_mult_distrib2
theorem sum_of_squares:
"6 * (\<Sum>i::nat=0..n. i^Suc (Suc 0)) = n * (n + 1) * (2 * n + 1)"
(is "?P n" is "?S n = _")
proof (induct n)
show "?P 0" by simp
next
fix n
have "?S (n + 1) = ?S n + 6 * (n + 1)^Suc (Suc 0)"
by (simp add: distrib)
also assume "?S n = n * (n + 1) * (2 * n + 1)"
also have "\<dots> + 6 * (n + 1)^Suc (Suc 0) =
(n + 1) * (n + 2) * (2 * (n + 1) + 1)"
by (simp add: distrib)
finally show "?P (Suc n)"
by simp
qed
theorem sum_of_cubes:
"4 * (\<Sum>i::nat=0..n. i^3) = (n * (n + 1))^Suc (Suc 0)"
(is "?P n" is "?S n = _")
proof (induct n)
show "?P 0" by (simp add: power_eq_if)
next
fix n
have "?S (n + 1) = ?S n + 4 * (n + 1)^3"
by (simp add: power_eq_if distrib)
also assume "?S n = (n * (n + 1))^Suc (Suc 0)"
also have "\<dots> + 4 * (n + 1)^3 = ((n + 1) * ((n + 1) + 1))^Suc (Suc 0)"
by (simp add: power_eq_if distrib)
finally show "?P (Suc n)"
by simp
qed
text \<open>Note that in contrast to older traditions of tactical proof
scripts, the structured proof applies induction on the original,
unsimplified statement. This allows to state the induction cases robustly
and conveniently. Simplification (or other automated) methods are then
applied in terminal position to solve certain sub-problems completely.
As a general rule of good proof style, automatic methods such as \<open>simp\<close> or
\<open>auto\<close> should normally be never used as initial proof methods with a
nested sub-proof to address the automatically produced situation, but only
as terminal ones to solve sub-problems.\<close>
end