(* Title: HOL/Isar_Examples/Summation.thy
Author: Makarius
*)
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 \<^theory_text>\<open>also\<close>~/ \<^theory_text>\<open>finally\<close> calculational chains may be just
anything. There is nothing special about \<^theory_text>\<open>have\<close>, so the natural deduction
element \<^theory_text>\<open>assume\<close> works just as well.
\<^enum> There are two \<^emph>\<open>separate\<close> primitives for building natural deduction
contexts: \<^theory_text>\<open>fix x\<close> and \<^theory_text>\<open>assume 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