text\<open>
This theoryis separate from\<^theory>\<open>HOL-Library.Tree\<close> because the former is discrete and
builds on \<^theory>\<open>Main\<close> whereas this theory builds on \<^theory>\<open>Complex_Main\<close>. \<close>
lemma min_height_size1_log: "min_height t \ log 2 (size1 t)" by (simp add: le_log2_of_power min_height_size1)
lemma size1_log_if_complete: "complete t \ height t = log 2 (size1 t)" by (simp add: size1_if_complete)
lemma min_height_size1_log_if_incomplete: "\ complete t \ min_height t < log 2 (size1 t)" by (simp add: less_log2_of_power min_height_size1_if_incomplete)
lemma min_height_acomplete: assumes"acomplete t" shows"min_height t = nat(floor(log 2 (size1 t)))" proof cases assume *: "complete t" hence"size1 t = 2 ^ min_height t" by (simp add: complete_iff_height size1_if_complete) from log2_of_power_eq[OF this] show ?thesis by linarith next assume *: "\ complete t" hence"height t = min_height t + 1" using assms min_height_le_height[of t] by(auto simp: acomplete_def complete_iff_height) hence"size1 t < 2 ^ (min_height t + 1)"by (metis * size1_height_if_incomplete) from floor_log_nat_eq_if[OF min_height_size1 this] show ?thesis by simp qed
lemma height_acomplete: assumes"acomplete t" shows"height t = nat(ceiling(log 2 (size1 t)))" proof cases assume *: "complete t" hence"size1 t = 2 ^ height t"by (simp add: size1_if_complete) from log2_of_power_eq[OF this] show ?thesis by linarith next assume *: "\ complete t" hence **: "height t = min_height t + 1" using assms min_height_le_height[of t] by(auto simp add: acomplete_def complete_iff_height) hence"size1 t \ 2 ^ (min_height t + 1)" by (metis size1_height) from log2_of_power_le[OF this size1_ge0] min_height_size1_log_if_incomplete[OF *] ** show ?thesis by linarith qed
lemma acomplete_Node_if_wbal1: assumes"acomplete l""acomplete r""size l = size r + 1" shows"acomplete \l, x, r\" proof - from assms(3) have [simp]: "size1 l = size1 r + 1"by(simp add: size1_size) have"nat \log 2 (1 + size1 r)\ \ nat \log 2 (size1 r)\" by(rule nat_mono[OF ceiling_mono]) simp hence 1: "height(Node l x r) = nat \log 2 (1 + size1 r)\ + 1" using height_acomplete[OF assms(1)] height_acomplete[OF assms(2)] by (simp del: nat_ceiling_le_eq add: max_def) have"nat \log 2 (1 + size1 r)\ \ nat \log 2 (size1 r)\" by(rule nat_mono[OF floor_mono]) simp hence 2: "min_height(Node l x r) = nat \log 2 (size1 r)\ + 1" using min_height_acomplete[OF assms(1)] min_height_acomplete[OF assms(2)] by (simp) have"size1 r \ 1" by(simp add: size1_size) thenobtain i where i: "2 ^ i \ size1 r" "size1 r < 2 ^ (i + 1)" using ex_power_ivl1[of 2 "size1 r"] by auto hence i1: "2 ^ i < size1 r + 1""size1 r + 1 \ 2 ^ (i + 1)" by auto from 1 2 floor_log_nat_eq_if[OF i] ceiling_log_nat_eq_if[OF i1] show ?thesis by(simp add:acomplete_def) qed
lemma acomplete_Node_if_wbal2: assumes"acomplete l""acomplete r""abs(int(size l) - int(size r)) \ 1" shows"acomplete \l, x, r\" proof - have"size l = size r \ (size l = size r + 1 \ size r = size l + 1)" (is "?A \ ?B") using assms(3) by linarith thus ?thesis proof assume"?A" thus ?thesis using assms(1,2) apply(simp add: acomplete_def min_def max_def) by (metis assms(1,2) acomplete_optimal le_antisym le_less) next assume"?B" thus ?thesis by (meson assms(1,2) acomplete_sym acomplete_Node_if_wbal1) qed qed
lemma acomplete_if_wbalanced: "wbalanced t \ acomplete t" proof(induction t) case Leaf show ?caseby (simp add: acomplete_def) next case (Node l x r) thus ?caseby(simp add: acomplete_Node_if_wbal2) qed
end
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