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products/Sources/formale Sprachen/Isabelle/HOL/Isar_Examples/ (Beweissystem Isabelle Version 2025-1^©) Datei vom 16.11.2025 mit Größe 5 kB

Quelle Summation.thy Sprache: Isabelle

(*  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 * (\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 "\ + 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:
  "(\i::nat=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 "\ + 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 * (\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 "\ + 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 * (\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 "\ + 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

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