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+(* Title: HOL/Isar_examples/ThreeDivides.thy
+ ID: $Id$
+ Author: Benjamin Porter, 2005
+*)
+
+header {* Three Divides Theorem *}
+
+theory ThreeDivides
+imports Main LaTeXsugar
+begin
+
+section {* Abstract *}
+
+text {*
+The following document presents a proof of the Three Divides N theorem
+formalised in the Isabelle/Isar theorem proving system.
+
+{\em Theorem}: 3 divides n if and only if 3 divides the sum of all
+digits in n.
+
+{\em Informal Proof}:
+Take $n = \sum{n_j * 10^j}$ where $n_j$ is the $j$'th least
+significant digit of the decimal denotation of the number n and the
+sum ranges over all digits. Then $$ (n - \sum{n_j}) = \sum{n_j * (10^j
+- 1)} $$ We know $\forall j\; 3|(10^j - 1) $ and hence $3|LHS$,
+therefore $$\forall n\; 3|n \Longleftrightarrow 3|\sum{n_j}$$
+@{text "\<box>"}
+*}
+
+section {* Formal proof *}
+
+subsection {* Miscellaneous summation lemmas *}
+
+text {* If $a$ divides @{text "A x"} for all x then $a$ divides any
+sum over terms of the form @{text "(A x)*(P x)"} for arbitrary $P$. *}
+
+lemma div_sum:
+ fixes a::nat and n::nat
+ shows "\<forall>x. a dvd A x \<Longrightarrow> a dvd (\<Sum>x<n. A x * D x)"
+proof (induct n)
+ case 0 show ?case by simp
+next
+ case (Suc n)
+ from Suc
+ have "a dvd (A n * D n)" by (simp add: dvd_mult2)
+ with Suc
+ have "a dvd ((\<Sum>x<n. A x * D x) + (A n * D n))" by (simp add: dvd_add)
+ thus ?case by simp
+qed
+
+subsection {* Generalised Three Divides *}
+
+text {* This section solves a generalised form of the three divides
+problem. Here we show that for any sequence of numbers the theorem
+holds. In the next section we specialise this theorem to apply
+directly to the decimal expansion of the natural numbers. *}
+
+text {* Here we show that the first statement in the informal proof is
+true for all natural numbers. Note we are using @{term "D i"} to
+denote the $i$'th element in a sequence of numbers. *}
+
+lemma digit_diff_split:
+ fixes n::nat and nd::nat and x::nat
+ shows "\<And>n. n = (\<Sum>x\<in>{..<nd}. (D x)*((10::nat)^x)) \<Longrightarrow>
+ (n - (\<Sum>x<nd. (D x))) = (\<Sum>x<nd. (D x)*(10^x - 1))"
+by (simp add: sum_diff_distrib diff_mult_distrib2)
+
+text {* Now we prove that 3 always divides numbers of the form $10^x - 1$. *}
+lemma three_divs_0 [rule_format, simplified]:
+ shows "(3::nat) dvd (10^x - 1)"
+proof (induct x)
+ case 0 show ?case by simp
+next
+ case (Suc n)
+ let ?thr = "(3::nat)"
+ have "?thr dvd 9" by simp
+ moreover
+ have "?thr dvd (10*(10^n - 1))" by (rule dvd_mult)
+ hence "?thr dvd (10^(n+1) - 10)" by (simp add: nat_distrib)
+ ultimately
+ have"?thr dvd ((10^(n+1) - 10) + 9)"
+ by (simp only: add_ac) (rule dvd_add)
+ thus ?case by simp
+qed
+
+text {* Expanding on the previous lemma and lemma @{text "div_sum\<dots>"} *}
+lemma three_divs_1:
+ fixes D :: "nat \<Rightarrow> nat"
+ shows "3 dvd (\<Sum>x<nd. D x * (10^x - 1))"
+ by (subst nat_mult_commute, rule div_sum) (simp add: three_divs_0)
+
+text {* Using lemmas @{text "digit_diff_split"} and
+@{text "three_divs_1"} we now prove the following lemma.
+*}
+lemma three_divs_2:
+ fixes nd::nat and D::"nat\<Rightarrow>nat"
+ shows "3 dvd ((\<Sum>x<nd. (D x)*(10^x)) - (\<Sum>x<nd. (D x)))"
+proof (simp only: digit_diff_split)
+ from three_divs_1 show "3 dvd (\<Sum>x<nd. D x * (10 ^ x - 1))" .
+qed
+
+text {*
+We now present the final theorem of this section. For any
+sequence of numbers (defined by a function @{term "D :: (nat\<Rightarrow>nat)"}),
+we show that 3 divides the expansive sum $\sum{(D\;x)*10^x}$ over $x$
+if and only if 3 divides the sum of the individual numbers
+$\sum{D\;x}$.
+*}
+lemma three_div_general:
+ fixes D :: "nat \<Rightarrow> nat"
+ shows "(3 dvd (\<Sum>x<nd. D x * 10^x)) = (3 dvd (\<Sum>x<nd. D x))"
+proof
+ have mono: "(\<Sum>x<nd. D x) \<le> (\<Sum>x<nd. D x * 10^x)"
+ by (rule setsum_mono, simp)
+ txt {* This lets us form the term
+ @{term "(\<Sum>x<nd. D x * 10^x) - (\<Sum>x<nd. D x)"} *}
+
+ {
+ assume "3 dvd (\<Sum>x<nd. D x)"
+ with three_divs_2 mono
+ show "3 dvd (\<Sum>x<nd. D x * 10^x)"
+ by (blast intro: dvd_diffD)
+ }
+ {
+ assume "3 dvd (\<Sum>x<nd. D x * 10^x)"
+ with three_divs_2 mono
+ show "3 dvd (\<Sum>x<nd. D x)"
+ by (blast intro: dvd_diffD1)
+ }
+qed
+
+
+subsection {* Three Divides Natural *}
+
+text {* This section shows that for all natural numbers we can
+generate a sequence of digits less than ten that represent the decimal
+expansion of the number. We then use the lemma @{text
+"three_div_general"} to prove our final theorem. *}
+
+subsubsection {* Definitions of length and digit sum *}
+
+text {* This section introduces some functions to calculate the
+required properties of natural numbers. We then proceed to prove some
+properties of these functions.
+
+The function @{text "nlen"} returns the number of digits in a natural
+number n. *}
+
+consts nlen :: "nat \<Rightarrow> nat"
+recdef nlen "measure id"
+ "nlen 0 = 0"
+ "nlen x = 1 + nlen (x div 10)"
+
+text {* The function @{text "sumdig"} returns the sum of all digits in
+some number n. *}
+
+constdefs
+ sumdig :: "nat \<Rightarrow> nat"
+ "sumdig n \<equiv> \<Sum>x < nlen n. n div 10^x mod 10"
+
+text {* Some properties of these functions follow. *}
+
+lemma nlen_zero:
+ "0 = nlen x \<Longrightarrow> x = 0"
+ by (induct x rule: nlen.induct) auto
+
+lemma nlen_suc:
+ "Suc m = nlen n \<Longrightarrow> m = nlen (n div 10)"
+ by (induct n rule: nlen.induct) simp_all
+
+
+text {* The following lemma is the principle lemma required to prove
+our theorem. It states that an expansion of some natural number $n$
+into a sequence of its individual digits is always possible. *}
+
+lemma exp_exists:
+ "\<And>m. nd = nlen m \<Longrightarrow> m = (\<Sum>x<nd. (m div (10::nat)^x mod 10) * 10^x)"
+proof (induct nd)
+ case 0 thus ?case by (simp add: nlen_zero)
+next
+ case (Suc nd)
+ hence IH:
+ "nd = nlen (m div 10) \<Longrightarrow>
+ m div 10 = (\<Sum>x<nd. m div 10 div 10^x mod 10 * 10^x)"
+ by blast
+ have "\<exists>c. m = 10*(m div 10) + c \<and> c < 10" by presburger
+ from this obtain c where mexp: "m = 10*(m div 10) + c \<and> c < 10" ..
+ then have cdef: "c = m mod 10" by arith
+ show "m = (\<Sum>x<Suc nd. m div 10^x mod 10 * 10^x)"
+ proof -
+ have "Suc nd = nlen m" .
+ then have
+ "nd = nlen (m div 10)" by (rule nlen_suc)
+ with IH have
+ "m div 10 = (\<Sum>x<nd. m div 10 div 10^x mod 10 * 10^x)" by simp
+ with mexp have
+ "m = 10*(\<Sum>x<nd. m div 10 div 10^x mod 10 * 10^x) + c" by simp
+ also have
+ "\<dots> = (\<Sum>x<nd. m div 10 div 10^x mod 10 * 10^(x+1)) + c"
+ by (subst setsum_mult) (simp add: mult_ac)
+ also have
+ "\<dots> = (\<Sum>x<nd. m div 10^(Suc x) mod 10 * 10^(Suc x)) + c"
+ by (simp add: div_mult2_eq[symmetric])
+ also have
+ "\<dots> = (\<Sum>x\<in>{Suc 0..<Suc nd}. m div 10^x mod 10 * 10^x) + c"
+ by (simp only: setsum_shift_bounds_Suc_ivl)
+ (simp add: atLeast0LessThan)
+ also have
+ "\<dots> = (\<Sum>x<Suc nd. m div 10^x mod 10 * 10^x)"
+ by (simp add: setsum_rmv_head [symmetric] cdef)
+ finally
+ show "m = (\<Sum>x<Suc nd. m div 10^x mod 10 * 10^x)" .
+ qed
+qed
+
+
+subsubsection {* Final theorem *}
+
+text {* We now combine the general theorem @{text "three_div_general"}
+and existence result of @{text "exp_exists"} to prove our final
+theorem. *}
+
+theorem three_divides_nat:
+ shows "(3 dvd n) = (3 dvd sumdig n)"
+proof (unfold sumdig_def)
+ obtain nd where "nd = nlen n" by simp
+ moreover
+ have "n = (\<Sum>x<nd. (n div (10::nat)^x mod 10) * 10^x)"
+ by (rule exp_exists)
+ moreover
+ have "3 dvd (\<Sum>x<nlen n. (n div (10::nat)^x mod 10) * 10^x) =
+ (3 dvd (\<Sum>x<nlen n. n div 10^x mod 10))"
+ by (rule three_div_general)
+ ultimately
+ show "3 dvd n = (3 dvd (\<Sum>x<nlen n. n div 10^x mod 10))" by simp
+qed
+
+
+end