src/HOL/Isar_examples/Summation.thy

 primrec
   "sum f 0 = 0"
   "sum f (Suc n) = f n + sum f n";
 
 syntax
-  "_SUM" :: "idt => nat => nat => nat"  ("SUM _ < _. _" [0, 0, 10] 10);
+  "_SUM" :: "idt => nat => nat => nat"
+    ("SUM _ < _. _" [0, 0, 10] 10);
 translations
   "SUM i < k. b" == "sum (%i. b) k";
 
 
 subsection {* Summation laws *};
+
+verbatim {* \begin{comment} *};
 
 (* FIXME binary arithmetic does not yet work here *)
 
 syntax
   "3" :: nat  ("3")

   "4" == "Suc 3"
   "6" == "Suc (Suc 4)";
 
 theorems [simp] = add_mult_distrib add_mult_distrib2 mult_ac;
 
+verbatim {* \end{comment} *};
 
-theorem sum_of_naturals: "2 * (SUM i < n + 1. i) = n * (n + 1)"
+
+theorem sum_of_naturals:
+  "2 * (SUM i < n + 1. i) = n * (n + 1)"
   (is "?P n" is "?S n = _");
 proof (induct n);
   show "?P 0"; by simp;
 
   fix n;

   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;
 
-theorem sum_of_odds: "(SUM i < n. 2 * i + 1) = n^2"
+theorem sum_of_odds:
+  "(SUM i < n. 2 * i + 1) = n^2"
   (is "?P n" is "?S n = _");
 proof (induct n);
   show "?P 0"; by simp;
 
   fix n;

   show "?P 0"; by simp;
 
   fix n;
   have "?S (n + 1) = ?S n + 6 * (n + 1)^2"; by simp;
   also; assume "?S n = n * (n + 1) * (2 * n + 1)";
-  also; have "... + 6 * (n + 1)^2 = (n + 1) * (n + 2) * (2 * (n + 1) + 1)";
-    by simp;
+  also; have "... + 6 * (n + 1)^2 =
+    (n + 1) * (n + 2) * (2 * (n + 1) + 1)"; by simp;
   finally; show "?P (Suc n)"; by simp;
 qed;
 
-theorem sum_of_cubes: "4 * (SUM i < n + 1. i^3) = (n * (n + 1))^2"
+theorem sum_of_cubes:
+  "4 * (SUM i < n + 1. i^3) = (n * (n + 1))^2"
   (is "?P n" is "?S n = _");
 proof (induct n);
   show "?P 0"; by simp;
 
   fix n;