src/HOL/Isar_examples/Summation.thy
 author wenzelm Sat Oct 30 20:20:48 1999 +0200 (1999-10-30) changeset 7982 d534b897ce39 parent 7968 964b65b4e433 child 8584 016314c2fa0a permissions -rw-r--r--
improved presentation;
     1 (*  Title:      HOL/Isar_examples/Summation.thy

     2     ID:         $Id$

     3     Author:     Markus Wenzel

     4 *)

     5

     6 header {* Summing natural numbers *};

     7

     8 theory Summation = Main:;

     9

    10 text_raw {*

    11  \footnote{This example is somewhat reminiscent of the

    12  \url{http://isabelle.in.tum.de/library/HOL/ex/NatSum.html}, which is

    13  discussed in \cite{isabelle-ref} in the context of permutative

    14  rewrite rules and ordered rewriting.}

    15 *};

    16

    17 text {*

    18  Subsequently, we prove some summation laws of natural numbers

    19  (including odds, squares, and cubes).  These examples demonstrate how

    20  plain natural deduction (including induction) may be combined with

    21  calculational proof.

    22 *};

    23

    24

    25 subsection {* A summation operator *};

    26

    27 text {*

    28   The binder operator $\idt{sum} :: (\idt{nat} \To \idt{nat}) \To   29 \idt{nat} \To \idt{nat}$ formalizes summation of natural numbers

    30  indexed from $0$ up to $k$ (excluding the bound):

    31  $  32 \sum\limits_{i < k} f(i) = \idt{sum} \ap (\lam i f \ap i) \ap k   33$

    34 *};

    35

    36 consts

    37   sum :: "[nat => nat, nat] => nat";

    38

    39 primrec

    40   "sum f 0 = 0"

    41   "sum f (Suc n) = f n + sum f n";

    42

    43 syntax

    44   "_SUM" :: "[idt, nat, nat] => nat"

    45     ("SUM _ < _. _" [0, 0, 10] 10);

    46 translations

    47   "SUM i < k. b" == "sum (\<lambda>i. b) k";

    48

    49

    50 subsection {* Summation laws *};

    51

    52 text_raw {* \begin{comment} *};

    53

    54 (* FIXME binary arithmetic does not yet work here *)

    55

    56 syntax

    57   "3" :: nat  ("3")

    58   "4" :: nat  ("4")

    59   "6" :: nat  ("6");

    60

    61 translations

    62   "3" == "Suc 2"

    63   "4" == "Suc 3"

    64   "6" == "Suc (Suc 4)";

    65

    66 theorems [simp] = add_mult_distrib add_mult_distrib2 mult_ac;

    67

    68 text_raw {* \end{comment} *};

    69

    70 text {*

    71  The sum of natural numbers $0 + \cdots + n$ equals $n \times (n +   72 1)/2$.  Avoiding formal reasoning about division we prove this

    73  equation multiplied by $2$.

    74 *};

    75

    76 theorem sum_of_naturals:

    77   "2 * (SUM i < n + 1. i) = n * (n + 1)"

    78   (is "?P n" is "?S n = _");

    79 proof (induct n);

    80   show "?P 0"; by simp;

    81

    82   fix n;

    83   have "?S (n + 1) = ?S n + 2 * (n + 1)"; by simp;

    84   also; assume "?S n = n * (n + 1)";

    85   also; have "... + 2 * (n + 1) = (n + 1) * (n + 2)"; by simp;

    86   finally; show "?P (Suc n)"; by simp;

    87 qed;

    88

    89 text {*

    90  The above proof is a typical instance of mathematical induction.  The

    91  main statement is viewed as some $\var{P} \ap n$ that is split by the

    92  induction method into base case $\var{P} \ap 0$, and step case

    93  $\var{P} \ap n \Impl \var{P} \ap (\idt{Suc} \ap n)$ for arbitrary $n$.

    94

    95  The step case is established by a short calculation in forward

    96  manner.  Starting from the left-hand side $\var{S} \ap (n + 1)$ of

    97  the thesis, the final result is achieved by transformations involving

    98  basic arithmetic reasoning (using the Simplifier).  The main point is

    99  where the induction hypothesis $\var{S} \ap n = n \times (n + 1)$ is

   100  introduced in order to replace a certain subterm.  So the

   101  transitivity'' rule involved here is actual \emph{substitution}.

   102  Also note how the occurrence of \dots'' in the subsequent step

   103  documents the position where the right-hand side of the hypothesis

   104  got filled in.

   105

   106  \medskip A further notable point here is integration of calculations

   107  with plain natural deduction.  This works so well in Isar for two

   108  reasons.

   109  \begin{enumerate}

   110

   111  \item Facts involved in \isakeyword{also}~/ \isakeyword{finally}

   112  calculational chains may be just anything.  There is nothing special

   113  about \isakeyword{have}, so the natural deduction element

   114  \isakeyword{assume} works just as well.

   115

   116  \item There are two \emph{separate} primitives for building natural

   117  deduction contexts: \isakeyword{fix}~$x$ and \isakeyword{assume}~$A$.

   118  Thus it is possible to start reasoning with some new arbitrary, but

   119  fixed'' elements before bringing in the actual assumption.  In

   120  contrast, natural deduction is occasionally formalized with basic

   121  context elements of the form $x:A$ instead.

   122

   123  \end{enumerate}

   124 *};

   125

   126 text {*

   127  \medskip We derive further summation laws for odds, squares, and

   128  cubes as follows.  The basic technique of induction plus calculation

   129  is the same as before.

   130 *};

   131

   132 theorem sum_of_odds:

   133   "(SUM i < n. 2 * i + 1) = n^2"

   134   (is "?P n" is "?S n = _");

   135 proof (induct n);

   136   show "?P 0"; by simp;

   137

   138   fix n;

   139   have "?S (n + 1) = ?S n + 2 * n + 1"; by simp;

   140   also; assume "?S n = n^2";

   141   also; have "... + 2 * n + 1 = (n + 1)^2"; by simp;

   142   finally; show "?P (Suc n)"; by simp;

   143 qed;

   144

   145 theorem sum_of_squares:

   146   "6 * (SUM i < n + 1. i^2) = n * (n + 1) * (2 * n + 1)"

   147   (is "?P n" is "?S n = _");

   148 proof (induct n);

   149   show "?P 0"; by simp;

   150

   151   fix n;

   152   have "?S (n + 1) = ?S n + 6 * (n + 1)^2"; by simp;

   153   also; assume "?S n = n * (n + 1) * (2 * n + 1)";

   154   also; have "... + 6 * (n + 1)^2 =

   155     (n + 1) * (n + 2) * (2 * (n + 1) + 1)"; by simp;

   156   finally; show "?P (Suc n)"; by simp;

   157 qed;

   158

   159 theorem sum_of_cubes:

   160   "4 * (SUM i < n + 1. i^3) = (n * (n + 1))^2"

   161   (is "?P n" is "?S n = _");

   162 proof (induct n);

   163   show "?P 0"; by simp;

   164

   165   fix n;

   166   have "?S (n + 1) = ?S n + 4 * (n + 1)^3"; by simp;

   167   also; assume "?S n = (n * (n + 1))^2";

   168   also; have "... + 4 * (n + 1)^3 = ((n + 1) * ((n + 1) + 1))^2";

   169     by simp;

   170   finally; show "?P (Suc n)"; by simp;

   171 qed;

   172

   173 text {*

   174  Comparing these examples with the tactic script version

   175  \url{http://isabelle.in.tum.de/library/HOL/ex/NatSum.html}, we note

   176  an important difference of how induction vs.\ simplification is

   177  applied.  While \cite[\S10]{isabelle-ref} advises for these examples

   178  that induction should not be applied until the goal is in the

   179  simplest form'' this would be a very bad idea in our setting.

   180

   181  Simplification normalizes all arithmetic expressions involved,

   182  producing huge intermediate goals.  With applying induction

   183  afterwards, the Isar proof text would have to reflect the emerging

   184  configuration by appropriate sub-proofs.  This would result in badly

   185  structured, low-level technical reasoning, without any good idea of

   186  the actual point.

   187

   188  \medskip As a general rule of good proof style, automatic methods

   189  such as $\idt{simp}$ or $\idt{auto}$ should normally be never used as

   190  initial proof methods, but only as terminal ones, solving certain

   191  goals completely.

   192 *};

   193

   194 end;