src/HOL/Isar_Examples/Summation.thy

recovered from a9c0572109af;

(*  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