# HG changeset patch
# User wenzelm
# Date 1485791240 -3600
# Node ID d0e55f85fd8a1b42a61d907b9a3c9a7b5e445289
# Parent  ea56dd12deb0d83b58509ab6729aab6ba5ed5d21
misc tuning and modernization;

diff -r ea56dd12deb0 -r d0e55f85fd8a src/HOL/ex/NatSum.thy
--- a/src/HOL/ex/NatSum.thy	Wed Feb 01 13:50:20 2017 +0100
+++ b/src/HOL/ex/NatSum.thy	Mon Jan 30 16:47:20 2017 +0100
@@ -1,124 +1,120 @@
-(*  Title:  HOL/ex/NatSum.thy
-    Author: Tobias Nipkow
+(*  Title:      HOL/ex/NatSum.thy
+    Author:     Tobias Nipkow
 *)
 
 section \<open>Summing natural numbers\<close>
 
-theory NatSum imports Main begin
+theory NatSum
+  imports Main
+begin
 
 text \<open>
   Summing natural numbers, squares, cubes, etc.
 
   Thanks to Sloane's On-Line Encyclopedia of Integer Sequences,
-  \<^url>\<open>http://www.research.att.com/~njas/sequences\<close>.
+  \<^url>\<open>http://oeis.org\<close>.
 \<close>
 
 lemmas [simp] =
   ring_distribs
   diff_mult_distrib diff_mult_distrib2 \<comment>\<open>for type nat\<close>
 
-text \<open>
-  \medskip The sum of the first \<open>n\<close> odd numbers equals \<open>n\<close>
-  squared.
-\<close>
+
+text \<open>\<^medskip> The sum of the first \<open>n\<close> odd numbers equals \<open>n\<close> squared.\<close>
 
 lemma sum_of_odds: "(\<Sum>i=0..<n. Suc (i + i)) = n * n"
   by (induct n) auto
 
 
-text \<open>
-  \medskip The sum of the first \<open>n\<close> odd squares.
-\<close>
+text \<open>\<^medskip> The sum of the first \<open>n\<close> odd squares.\<close>
 
 lemma sum_of_odd_squares:
   "3 * (\<Sum>i=0..<n. Suc(2*i) * Suc(2*i)) = n * (4 * n * n - 1)"
   by (induct n) auto
 
 
-text \<open>
-  \medskip The sum of the first \<open>n\<close> odd cubes
-\<close>
+text \<open>\<^medskip> The sum of the first \<open>n\<close> odd cubes.\<close>
 
 lemma sum_of_odd_cubes:
   "(\<Sum>i=0..<n. Suc (2*i) * Suc (2*i) * Suc (2*i)) =
     n * n * (2 * n * n - 1)"
   by (induct n) auto
 
-text \<open>
-  \medskip The sum of the first \<open>n\<close> positive integers equals
-  \<open>n (n + 1) / 2\<close>.\<close>
+
+text \<open>\<^medskip> The sum of the first \<open>n\<close> positive integers equals \<open>n (n + 1) / 2\<close>.\<close>
 
-lemma sum_of_naturals:
-    "2 * (\<Sum>i=0..n. i) = n * Suc n"
+lemma sum_of_naturals: "2 * (\<Sum>i=0..n. i) = n * Suc n"
+  by (induct n) auto
+
+lemma sum_of_squares: "6 * (\<Sum>i=0..n. i * i) = n * Suc n * Suc (2 * n)"
   by (induct n) auto
 
-lemma sum_of_squares:
-    "6 * (\<Sum>i=0..n. i * i) = n * Suc n * Suc (2 * n)"
+lemma sum_of_cubes: "4 * (\<Sum>i=0..n. i * i * i) = n * n * Suc n * Suc n"
   by (induct n) auto
 
-lemma sum_of_cubes:
-    "4 * (\<Sum>i=0..n. i * i * i) = n * n * Suc n * Suc n"
-  by (induct n) auto
+
+text \<open>\<^medskip> A cute identity:\<close>
 
-text\<open>\medskip A cute identity:\<close>
-
-lemma sum_squared: "(\<Sum>i=0..n. i)^2 = (\<Sum>i=0..n::nat. i^3)"
-proof(induct n)
-  case 0 show ?case by simp
+lemma sum_squared: "(\<Sum>i=0..n. i)^2 = (\<Sum>i=0..n. i^3)" for n :: nat
+proof (induct n)
+  case 0
+  show ?case by simp
 next
   case (Suc n)
   have "(\<Sum>i = 0..Suc n. i)^2 =
         (\<Sum>i = 0..n. i^3) + (2*(\<Sum>i = 0..n. i)*(n+1) + (n+1)^2)"
     (is "_ = ?A + ?B")
-    using Suc by(simp add:eval_nat_numeral)
+    using Suc by (simp add: eval_nat_numeral)
   also have "?B = (n+1)^3"
-    using sum_of_naturals by(simp add:eval_nat_numeral)
+    using sum_of_naturals by (simp add: eval_nat_numeral)
   also have "?A + (n+1)^3 = (\<Sum>i=0..Suc n. i^3)" by simp
   finally show ?case .
 qed
 
-text \<open>
-  \medskip Sum of fourth powers: three versions.
-\<close>
+
+text \<open>\<^medskip> Sum of fourth powers: three versions.\<close>
 
 lemma sum_of_fourth_powers:
   "30 * (\<Sum>i=0..n. i * i * i * i) =
     n * Suc n * Suc (2 * n) * (3 * n * n + 3 * n - 1)"
-  apply (induct n)
-   apply simp_all
-  apply (case_tac n)  \<comment> \<open>eliminates the subtraction\<close> 
-   apply (simp_all (no_asm_simp))
-  done
+proof (induct n)
+  case 0
+  show ?case by simp
+next
+  case (Suc n)
+  then show ?case
+    by (cases n)  \<comment> \<open>eliminates the subtraction\<close>
+      simp_all
+qed
 
 text \<open>
   Two alternative proofs, with a change of variables and much more
-  subtraction, performed using the integers.\<close>
+  subtraction, performed using the integers.
+\<close>
 
 lemma int_sum_of_fourth_powers:
   "30 * int (\<Sum>i=0..<m. i * i * i * i) =
     int m * (int m - 1) * (int(2 * m) - 1) *
     (int(3 * m * m) - int(3 * m) - 1)"
-  by (induct m) (simp_all add: of_nat_mult)
+  by (induct m) simp_all
 
 lemma of_nat_sum_of_fourth_powers:
   "30 * of_nat (\<Sum>i=0..<m. i * i * i * i) =
     of_nat m * (of_nat m - 1) * (of_nat (2 * m) - 1) *
     (of_nat (3 * m * m) - of_nat (3 * m) - (1::int))"
-  by (induct m) (simp_all add: of_nat_mult)
+  by (induct m) simp_all
 
 
-text \<open>
-  \medskip Sums of geometric series: \<open>2\<close>, \<open>3\<close> and the
-  general case.
-\<close>
+text \<open>\<^medskip> Sums of geometric series: \<open>2\<close>, \<open>3\<close> and the general case.\<close>
 
 lemma sum_of_2_powers: "(\<Sum>i=0..<n. 2^i) = 2^n - (1::nat)"
-  by (induct n) (auto split: nat_diff_split)
+  by (induct n) auto
 
 lemma sum_of_3_powers: "2 * (\<Sum>i=0..<n. 3^i) = 3^n - (1::nat)"
   by (induct n) auto
 
-lemma sum_of_powers: "0 < k ==> (k - 1) * (\<Sum>i=0..<n. k^i) = k^n - (1::nat)"
+lemma sum_of_powers: "0 < k \<Longrightarrow> (k - 1) * (\<Sum>i=0..<n. k^i) = k^n - 1"
+  for k :: nat
   by (induct n) auto
 
 end