author nipkow Mon, 23 Sep 2019 17:15:29 +0200 changeset 70747 548420d389ea parent 70745 be8e617b6eb3 child 70748 b3b84b71e398
Enforced precodition "n <= length xs" to avoid relying on "hd []".
```--- a/src/HOL/Data_Structures/Balance.thy	Mon Sep 23 08:43:52 2019 +0200
+++ b/src/HOL/Data_Structures/Balance.thy	Mon Sep 23 17:15:29 2019 +0200
@@ -38,43 +38,32 @@
in (Node l (hd ys) r, zs))"
by(simp_all add: bal.simps)

-text\<open>Some of the following lemmas take advantage of the fact
-that \<open>bal xs n\<close> yields a result even if \<open>n > length xs\<close>.\<close>
-
-lemma size_bal: "bal n xs = (t,ys) \<Longrightarrow> size t = n"
-proof(induction n xs arbitrary: t ys rule: bal.induct)
-  case (1 n xs)
-  thus ?case
-    by(cases "n=0")
-      (auto simp add: bal_simps Let_def split: prod.splits)
-qed
-
lemma bal_inorder:
-  "\<lbrakk> bal n xs = (t,ys); n \<le> length xs \<rbrakk>
-  \<Longrightarrow> inorder t = take n xs \<and> ys = drop n xs"
-proof(induction n xs arbitrary: t ys rule: bal.induct)
+  "\<lbrakk> n \<le> length xs; bal n xs = (t,zs) \<rbrakk>
+  \<Longrightarrow> inorder t = take n xs \<and> zs = drop n xs"
+proof(induction n xs arbitrary: t zs rule: bal.induct)
case (1 n xs) show ?case
proof cases
assume "n = 0" thus ?thesis using 1 by (simp add: bal_simps)
next
assume [arith]: "n \<noteq> 0"
-    let ?n1 = "n div 2" let ?n2 = "n - 1 - ?n1"
-    from "1.prems" obtain l r xs' where
-      b1: "bal ?n1 xs = (l,xs')" and
-      b2: "bal ?n2 (tl xs') = (r,ys)" and
-      t: "t = \<langle>l, hd xs', r\<rangle>"
+    let ?m = "n div 2" let ?m' = "n - 1 - ?m"
+    from "1.prems"(2) obtain l r ys where
+      b1: "bal ?m xs = (l,ys)" and
+      b2: "bal ?m' (tl ys) = (r,zs)" and
+      t: "t = \<langle>l, hd ys, r\<rangle>"
by(auto simp: Let_def bal_simps split: prod.splits)
-    have IH1: "inorder l = take ?n1 xs \<and> xs' = drop ?n1 xs"
-      using b1 "1.prems" by(intro "1.IH"(1)) auto
-    have IH2: "inorder r = take ?n2 (tl xs') \<and> ys = drop ?n2 (tl xs')"
-      using b1 b2 IH1 "1.prems" by(intro "1.IH"(2)) auto
-    have "drop (n div 2) xs \<noteq> []" using "1.prems"(2) by simp
-    hence "hd (drop ?n1 xs) # take ?n2 (tl (drop ?n1 xs)) = take (?n2 + 1) (drop ?n1 xs)"
+    have IH1: "inorder l = take ?m xs \<and> ys = drop ?m xs"
+      using b1 "1.prems"(1) by(intro "1.IH"(1)) auto
+    have IH2: "inorder r = take ?m' (tl ys) \<and> zs = drop ?m' (tl ys)"
+      using b1 b2 IH1 "1.prems"(1) by(intro "1.IH"(2)) auto
+    have "drop (n div 2) xs \<noteq> []" using "1.prems"(1) by simp
+    hence "hd (drop ?m xs) # take ?m' (tl (drop ?m xs)) = take (?m' + 1) (drop ?m xs)"
by (metis Suc_eq_plus1 take_Suc)
hence *: "inorder t = take n xs" using t IH1 IH2
-      using take_add[of ?n1 "?n2+1" xs] by(simp)
+      using take_add[of ?m "?m'+1" xs] by(simp)
have "n - n div 2 + n div 2 = n" by simp
-    hence "ys = drop n xs" using IH1 IH2 by (simp add: drop_Suc[symmetric])
+    hence "zs = drop n xs" using IH1 IH2 by (simp add: drop_Suc[symmetric])
thus ?thesis using * by blast
qed
qed
@@ -93,41 +82,56 @@
corollary inorder_balance_tree[simp]: "inorder(balance_tree t) = inorder t"
by(simp add: balance_tree_def inorder_bal_tree)

-corollary size_bal_list[simp]: "size(bal_list n xs) = n"
+
+text\<open>The size lemmas below do not require the precondition @{prop"n \<le> length xs"}
+(or  @{prop"n \<le> size t"}) that they come with. They could take advantage of the fact
+that @{term "bal xs n"} yields a result even if @{prop "n > length xs"}.
+In that case the result will contain one or more occurrences of @{term "hd []"}.
+However, this is counter-intuitive and does not reflect the execution
+in an eager functional language.\<close>
+
+lemma size_bal: "\<lbrakk> n \<le> length xs; bal n xs = (t,zs) \<rbrakk> \<Longrightarrow> size t = n \<and> length zs = length xs - n"
+by (metis bal_inorder length_drop length_inorder length_take min.absorb2)
+
+corollary size_bal_list[simp]: "n \<le> length xs \<Longrightarrow> size(bal_list n xs) = n"
unfolding bal_list_def by (metis prod.collapse size_bal)

corollary size_balance_list[simp]: "size(balance_list xs) = length xs"
by (simp add: balance_list_def)

-corollary size_bal_tree[simp]: "size(bal_tree n t) = n"
+corollary size_bal_tree[simp]: "n \<le> size t \<Longrightarrow> size(bal_tree n t) = n"
by(simp add: bal_tree_def)

corollary size_balance_tree[simp]: "size(balance_tree t) = size t"
by(simp add: balance_tree_def)

+lemma pre_rec2: "\<lbrakk> n \<le> length xs; bal (n div 2) xs = (l, ys) \<rbrakk>
+ \<Longrightarrow> (n - 1 - n div 2) \<le> length(tl ys)"
+using size_bal[of "n div 2" xs l ys] by simp
+
lemma min_height_bal:
-  "bal n xs = (t,ys) \<Longrightarrow> min_height t = nat(\<lfloor>log 2 (n + 1)\<rfloor>)"
-proof(induction n xs arbitrary: t ys rule: bal.induct)
-  case (1 n xs) show ?case
+  "\<lbrakk> n \<le> length xs; bal n xs = (t,zs) \<rbrakk> \<Longrightarrow> min_height t = nat(\<lfloor>log 2 (n + 1)\<rfloor>)"
+proof(induction n xs arbitrary: t zs rule: bal.induct)
+  case (1 n xs)
+  show ?case
proof cases
-    assume "n = 0" thus ?thesis
-      using "1.prems" by (simp add: bal_simps)
+    assume "n = 0" thus ?thesis using "1.prems"(2) by (simp add: bal_simps)
next
assume [arith]: "n \<noteq> 0"
-    from "1.prems" obtain l r xs' where
-      b1: "bal (n div 2) xs = (l,xs')" and
-      b2: "bal (n - 1 - n div 2) (tl xs') = (r,ys)" and
-      t: "t = \<langle>l, hd xs', r\<rangle>"
+    from "1.prems" obtain l r ys where
+      b1: "bal (n div 2) xs = (l,ys)" and
+      b2: "bal (n - 1 - n div 2) (tl ys) = (r,zs)" and
+      t: "t = \<langle>l, hd ys, r\<rangle>"
by(auto simp: bal_simps Let_def split: prod.splits)
let ?log1 = "nat (floor(log 2 (n div 2 + 1)))"
let ?log2 = "nat (floor(log 2 (n - 1 - n div 2 + 1)))"
-    have IH1: "min_height l = ?log1" using "1.IH"(1) b1 by simp
-    have IH2: "min_height r = ?log2" using "1.IH"(2) b1 b2 by simp
+    have IH1: "min_height l = ?log1" using "1.IH"(1) b1 "1.prems"(1) by simp
+    have IH2: "min_height r = ?log2"
+      using "1.prems"(1) size_bal[OF _ b1] size_bal[OF _ b2] b1 b2 by(intro "1.IH"(2)) auto
have "(n+1) div 2 \<ge> 1" by arith
hence 0: "log 2 ((n+1) div 2) \<ge> 0" by simp
have "n - 1 - n div 2 + 1 \<le> n div 2 + 1" by arith
-    hence le: "?log2 \<le> ?log1"
-      by(simp add: nat_mono floor_mono)
+    hence le: "?log2 \<le> ?log1" by(simp add: nat_mono floor_mono)
have "min_height t = min ?log1 ?log2 + 1" by (simp add: t IH1 IH2)
also have "\<dots> = ?log2 + 1" using le by (simp add: min_absorb2)
also have "n - 1 - n div 2 + 1 = (n+1) div 2" by linarith
@@ -141,24 +145,27 @@
qed

lemma height_bal:
-  "bal n xs = (t,ys) \<Longrightarrow> height t = nat \<lceil>log 2 (n + 1)\<rceil>"
-proof(induction n xs arbitrary: t ys rule: bal.induct)
+  "\<lbrakk> n \<le> length xs; bal n xs = (t,zs) \<rbrakk> \<Longrightarrow> height t = nat \<lceil>log 2 (n + 1)\<rceil>"
+proof(induction n xs arbitrary: t zs rule: bal.induct)
case (1 n xs) show ?case
proof cases
assume "n = 0" thus ?thesis
using "1.prems" by (simp add: bal_simps)
next
assume [arith]: "n \<noteq> 0"
-    from "1.prems" obtain l r xs' where
-      b1: "bal (n div 2) xs = (l,xs')" and
-      b2: "bal (n - 1 - n div 2) (tl xs') = (r,ys)" and
-      t: "t = \<langle>l, hd xs', r\<rangle>"
+    from "1.prems" obtain l r ys where
+      b1: "bal (n div 2) xs = (l,ys)" and
+      b2: "bal (n - 1 - n div 2) (tl ys) = (r,zs)" and
+      t: "t = \<langle>l, hd ys, r\<rangle>"
by(auto simp: bal_simps Let_def split: prod.splits)
let ?log1 = "nat \<lceil>log 2 (n div 2 + 1)\<rceil>"
let ?log2 = "nat \<lceil>log 2 (n - 1 - n div 2 + 1)\<rceil>"
-    have IH1: "height l = ?log1" using "1.IH"(1) b1 by simp
-    have IH2: "height r = ?log2" using "1.IH"(2) b1 b2 by simp
-    have 0: "log 2 (n div 2 + 1) \<ge> 0" by auto
+    have 1: "n div 2 \<le> length xs" using "1.prems"(1) by linarith
+    have 2: "n - 1 - n div 2 \<le> length (tl ys)" using "1.prems"(1) size_bal[OF 1 b1] by simp
+    have IH1: "height l = ?log1" using "1.IH"(1) b1 "1.prems"(1) by simp
+    have IH2: "height r = ?log2"
+      using b1 b2 size_bal[OF _ b1] size_bal[OF _ b2] "1.prems"(1) by(intro "1.IH"(2)) auto
+    have 0: "log 2 (n div 2 + 1) \<ge> 0" by simp
have "n - 1 - n div 2 + 1 \<le> n div 2 + 1" by arith
hence le: "?log2 \<le> ?log1"
by(simp add: nat_mono ceiling_mono del: nat_ceiling_le_eq)
@@ -172,7 +179,7 @@
qed

lemma balanced_bal:
-  assumes "bal n xs = (t,ys)" shows "balanced t"
+  assumes "n \<le> length xs" "bal n xs = (t,ys)" shows "balanced t"
unfolding balanced_def
using height_bal[OF assms] min_height_bal[OF assms]
by linarith
@@ -186,59 +193,59 @@
by (simp add: balance_list_def height_bal_list)

corollary height_bal_tree:
-  "n \<le> length xs \<Longrightarrow> height (bal_tree n t) = nat\<lceil>log 2 (n + 1)\<rceil>"
+  "n \<le> size t \<Longrightarrow> height (bal_tree n t) = nat\<lceil>log 2 (n + 1)\<rceil>"
unfolding bal_list_def bal_tree_def
-using height_bal prod.exhaust_sel by blast
+by (metis bal_list_def height_bal_list length_inorder)

corollary height_balance_tree:
"height (balance_tree t) = nat\<lceil>log 2 (size t + 1)\<rceil>"
by (simp add: bal_tree_def balance_tree_def height_bal_list)

-corollary balanced_bal_list[simp]: "balanced (bal_list n xs)"
+corollary balanced_bal_list[simp]: "n \<le> length xs \<Longrightarrow> balanced (bal_list n xs)"
unfolding bal_list_def by (metis  balanced_bal prod.collapse)

corollary balanced_balance_list[simp]: "balanced (balance_list xs)"
by (simp add: balance_list_def)

-corollary balanced_bal_tree[simp]: "balanced (bal_tree n t)"
+corollary balanced_bal_tree[simp]: "n \<le> size t \<Longrightarrow> balanced (bal_tree n t)"
by (simp add: bal_tree_def)

corollary balanced_balance_tree[simp]: "balanced (balance_tree t)"
by (simp add: balance_tree_def)

-lemma wbalanced_bal: "bal n xs = (t,ys) \<Longrightarrow> wbalanced t"
+lemma wbalanced_bal: "\<lbrakk> n \<le> length xs; bal n xs = (t,ys) \<rbrakk> \<Longrightarrow> wbalanced t"
proof(induction n xs arbitrary: t ys rule: bal.induct)
case (1 n xs)
show ?case
proof cases
assume "n = 0"
-    thus ?thesis
-      using "1.prems" by(simp add: bal_simps)
+    thus ?thesis using "1.prems"(2) by(simp add: bal_simps)
next
-    assume "n \<noteq> 0"
+    assume [arith]: "n \<noteq> 0"
with "1.prems" obtain l ys r zs where
rec1: "bal (n div 2) xs = (l, ys)" and
rec2: "bal (n - 1 - n div 2) (tl ys) = (r, zs)" and
t: "t = \<langle>l, hd ys, r\<rangle>"
by(auto simp add: bal_simps Let_def split: prod.splits)
-    have l: "wbalanced l" using "1.IH"(1)[OF \<open>n\<noteq>0\<close> refl rec1] .
-    have "wbalanced r" using "1.IH"(2)[OF \<open>n\<noteq>0\<close> refl rec1[symmetric] refl rec2] .
-    with l t size_bal[OF rec1] size_bal[OF rec2]
+    have l: "wbalanced l" using "1.IH"(1)[OF \<open>n\<noteq>0\<close> refl _ rec1] "1.prems"(1) by linarith
+    have "wbalanced r"
+      using rec1 rec2 pre_rec2[OF "1.prems"(1) rec1] by(intro "1.IH"(2)) auto
+    with l t size_bal[OF _ rec1] size_bal[OF _ rec2] "1.prems"(1)
show ?thesis by auto
qed
qed

text\<open>An alternative proof via @{thm balanced_if_wbalanced}:\<close>
-lemma "bal n xs = (t,ys) \<Longrightarrow> balanced t"
+lemma "\<lbrakk> n \<le> length xs; bal n xs = (t,ys) \<rbrakk> \<Longrightarrow> balanced t"
by(rule balanced_if_wbalanced[OF wbalanced_bal])

-lemma wbalanced_bal_list[simp]: "wbalanced (bal_list n xs)"
+lemma wbalanced_bal_list[simp]: "n \<le> length xs \<Longrightarrow> wbalanced (bal_list n xs)"
by(simp add: bal_list_def) (metis prod.collapse wbalanced_bal)

lemma wbalanced_balance_list[simp]: "wbalanced (balance_list xs)"
by(simp add: balance_list_def)

-lemma wbalanced_bal_tree[simp]: "wbalanced (bal_tree n t)"
+lemma wbalanced_bal_tree[simp]: "n \<le> size t \<Longrightarrow> wbalanced (bal_tree n t)"
by(simp add: bal_tree_def)

lemma wbalanced_balance_tree: "wbalanced (balance_tree t)"```