src/HOL/Metis_Examples/Big_O.thy
changeset 53015 a1119cf551e8
parent 51130 76d68444cd59
child 54230 b1d955791529
     1.1 --- a/src/HOL/Metis_Examples/Big_O.thy	Tue Aug 13 14:20:22 2013 +0200
     1.2 +++ b/src/HOL/Metis_Examples/Big_O.thy	Tue Aug 13 16:25:47 2013 +0200
     1.3 @@ -42,20 +42,20 @@
     1.4  proof -
     1.5    fix c :: 'a and x :: 'b
     1.6    assume A1: "\<forall>x. \<bar>h x\<bar> \<le> c * \<bar>f x\<bar>"
     1.7 -  have F1: "\<forall>x\<^isub>1\<Colon>'a\<Colon>linordered_idom. 0 \<le> \<bar>x\<^isub>1\<bar>" by (metis abs_ge_zero)
     1.8 -  have F2: "\<forall>x\<^isub>1\<Colon>'a\<Colon>linordered_idom. 1 * x\<^isub>1 = x\<^isub>1" by (metis mult_1)
     1.9 -  have F3: "\<forall>x\<^isub>1 x\<^isub>3. x\<^isub>3 \<le> \<bar>h x\<^isub>1\<bar> \<longrightarrow> x\<^isub>3 \<le> c * \<bar>f x\<^isub>1\<bar>" by (metis A1 order_trans)
    1.10 -  have F4: "\<forall>x\<^isub>2 x\<^isub>3\<Colon>'a\<Colon>linordered_idom. \<bar>x\<^isub>3\<bar> * \<bar>x\<^isub>2\<bar> = \<bar>x\<^isub>3 * x\<^isub>2\<bar>"
    1.11 +  have F1: "\<forall>x\<^sub>1\<Colon>'a\<Colon>linordered_idom. 0 \<le> \<bar>x\<^sub>1\<bar>" by (metis abs_ge_zero)
    1.12 +  have F2: "\<forall>x\<^sub>1\<Colon>'a\<Colon>linordered_idom. 1 * x\<^sub>1 = x\<^sub>1" by (metis mult_1)
    1.13 +  have F3: "\<forall>x\<^sub>1 x\<^sub>3. x\<^sub>3 \<le> \<bar>h x\<^sub>1\<bar> \<longrightarrow> x\<^sub>3 \<le> c * \<bar>f x\<^sub>1\<bar>" by (metis A1 order_trans)
    1.14 +  have F4: "\<forall>x\<^sub>2 x\<^sub>3\<Colon>'a\<Colon>linordered_idom. \<bar>x\<^sub>3\<bar> * \<bar>x\<^sub>2\<bar> = \<bar>x\<^sub>3 * x\<^sub>2\<bar>"
    1.15      by (metis abs_mult)
    1.16 -  have F5: "\<forall>x\<^isub>3 x\<^isub>1\<Colon>'a\<Colon>linordered_idom. 0 \<le> x\<^isub>1 \<longrightarrow> \<bar>x\<^isub>3 * x\<^isub>1\<bar> = \<bar>x\<^isub>3\<bar> * x\<^isub>1"
    1.17 +  have F5: "\<forall>x\<^sub>3 x\<^sub>1\<Colon>'a\<Colon>linordered_idom. 0 \<le> x\<^sub>1 \<longrightarrow> \<bar>x\<^sub>3 * x\<^sub>1\<bar> = \<bar>x\<^sub>3\<bar> * x\<^sub>1"
    1.18      by (metis abs_mult_pos)
    1.19 -  hence "\<forall>x\<^isub>1\<ge>0. \<bar>x\<^isub>1\<Colon>'a\<Colon>linordered_idom\<bar> = \<bar>1\<bar> * x\<^isub>1" by (metis F2)
    1.20 -  hence "\<forall>x\<^isub>1\<ge>0. \<bar>x\<^isub>1\<Colon>'a\<Colon>linordered_idom\<bar> = x\<^isub>1" by (metis F2 abs_one)
    1.21 -  hence "\<forall>x\<^isub>3. 0 \<le> \<bar>h x\<^isub>3\<bar> \<longrightarrow> \<bar>c * \<bar>f x\<^isub>3\<bar>\<bar> = c * \<bar>f x\<^isub>3\<bar>" by (metis F3)
    1.22 -  hence "\<forall>x\<^isub>3. \<bar>c * \<bar>f x\<^isub>3\<bar>\<bar> = c * \<bar>f x\<^isub>3\<bar>" by (metis F1)
    1.23 -  hence "\<forall>x\<^isub>3. (0\<Colon>'a) \<le> \<bar>f x\<^isub>3\<bar> \<longrightarrow> c * \<bar>f x\<^isub>3\<bar> = \<bar>c\<bar> * \<bar>f x\<^isub>3\<bar>" by (metis F5)
    1.24 -  hence "\<forall>x\<^isub>3. (0\<Colon>'a) \<le> \<bar>f x\<^isub>3\<bar> \<longrightarrow> c * \<bar>f x\<^isub>3\<bar> = \<bar>c * f x\<^isub>3\<bar>" by (metis F4)
    1.25 -  hence "\<forall>x\<^isub>3. c * \<bar>f x\<^isub>3\<bar> = \<bar>c * f x\<^isub>3\<bar>" by (metis F1)
    1.26 +  hence "\<forall>x\<^sub>1\<ge>0. \<bar>x\<^sub>1\<Colon>'a\<Colon>linordered_idom\<bar> = \<bar>1\<bar> * x\<^sub>1" by (metis F2)
    1.27 +  hence "\<forall>x\<^sub>1\<ge>0. \<bar>x\<^sub>1\<Colon>'a\<Colon>linordered_idom\<bar> = x\<^sub>1" by (metis F2 abs_one)
    1.28 +  hence "\<forall>x\<^sub>3. 0 \<le> \<bar>h x\<^sub>3\<bar> \<longrightarrow> \<bar>c * \<bar>f x\<^sub>3\<bar>\<bar> = c * \<bar>f x\<^sub>3\<bar>" by (metis F3)
    1.29 +  hence "\<forall>x\<^sub>3. \<bar>c * \<bar>f x\<^sub>3\<bar>\<bar> = c * \<bar>f x\<^sub>3\<bar>" by (metis F1)
    1.30 +  hence "\<forall>x\<^sub>3. (0\<Colon>'a) \<le> \<bar>f x\<^sub>3\<bar> \<longrightarrow> c * \<bar>f x\<^sub>3\<bar> = \<bar>c\<bar> * \<bar>f x\<^sub>3\<bar>" by (metis F5)
    1.31 +  hence "\<forall>x\<^sub>3. (0\<Colon>'a) \<le> \<bar>f x\<^sub>3\<bar> \<longrightarrow> c * \<bar>f x\<^sub>3\<bar> = \<bar>c * f x\<^sub>3\<bar>" by (metis F4)
    1.32 +  hence "\<forall>x\<^sub>3. c * \<bar>f x\<^sub>3\<bar> = \<bar>c * f x\<^sub>3\<bar>" by (metis F1)
    1.33    hence "\<bar>h x\<bar> \<le> \<bar>c * f x\<bar>" by (metis A1)
    1.34    thus "\<bar>h x\<bar> \<le> \<bar>c\<bar> * \<bar>f x\<bar>" by (metis F4)
    1.35  qed
    1.36 @@ -73,12 +73,12 @@
    1.37  proof -
    1.38    fix c :: 'a and x :: 'b
    1.39    assume A1: "\<forall>x. \<bar>h x\<bar> \<le> c * \<bar>f x\<bar>"
    1.40 -  have F1: "\<forall>x\<^isub>1\<Colon>'a\<Colon>linordered_idom. 1 * x\<^isub>1 = x\<^isub>1" by (metis mult_1)
    1.41 -  have F2: "\<forall>x\<^isub>2 x\<^isub>3\<Colon>'a\<Colon>linordered_idom. \<bar>x\<^isub>3\<bar> * \<bar>x\<^isub>2\<bar> = \<bar>x\<^isub>3 * x\<^isub>2\<bar>"
    1.42 +  have F1: "\<forall>x\<^sub>1\<Colon>'a\<Colon>linordered_idom. 1 * x\<^sub>1 = x\<^sub>1" by (metis mult_1)
    1.43 +  have F2: "\<forall>x\<^sub>2 x\<^sub>3\<Colon>'a\<Colon>linordered_idom. \<bar>x\<^sub>3\<bar> * \<bar>x\<^sub>2\<bar> = \<bar>x\<^sub>3 * x\<^sub>2\<bar>"
    1.44      by (metis abs_mult)
    1.45 -  have "\<forall>x\<^isub>1\<ge>0. \<bar>x\<^isub>1\<Colon>'a\<Colon>linordered_idom\<bar> = x\<^isub>1" by (metis F1 abs_mult_pos abs_one)
    1.46 -  hence "\<forall>x\<^isub>3. \<bar>c * \<bar>f x\<^isub>3\<bar>\<bar> = c * \<bar>f x\<^isub>3\<bar>" by (metis A1 abs_ge_zero order_trans)
    1.47 -  hence "\<forall>x\<^isub>3. 0 \<le> \<bar>f x\<^isub>3\<bar> \<longrightarrow> c * \<bar>f x\<^isub>3\<bar> = \<bar>c * f x\<^isub>3\<bar>" by (metis F2 abs_mult_pos)
    1.48 +  have "\<forall>x\<^sub>1\<ge>0. \<bar>x\<^sub>1\<Colon>'a\<Colon>linordered_idom\<bar> = x\<^sub>1" by (metis F1 abs_mult_pos abs_one)
    1.49 +  hence "\<forall>x\<^sub>3. \<bar>c * \<bar>f x\<^sub>3\<bar>\<bar> = c * \<bar>f x\<^sub>3\<bar>" by (metis A1 abs_ge_zero order_trans)
    1.50 +  hence "\<forall>x\<^sub>3. 0 \<le> \<bar>f x\<^sub>3\<bar> \<longrightarrow> c * \<bar>f x\<^sub>3\<bar> = \<bar>c * f x\<^sub>3\<bar>" by (metis F2 abs_mult_pos)
    1.51    hence "\<bar>h x\<bar> \<le> \<bar>c * f x\<bar>" by (metis A1 abs_ge_zero)
    1.52    thus "\<bar>h x\<bar> \<le> \<bar>c\<bar> * \<bar>f x\<bar>" by (metis F2)
    1.53  qed
    1.54 @@ -96,10 +96,10 @@
    1.55  proof -
    1.56    fix c :: 'a and x :: 'b
    1.57    assume A1: "\<forall>x. \<bar>h x\<bar> \<le> c * \<bar>f x\<bar>"
    1.58 -  have F1: "\<forall>x\<^isub>1\<Colon>'a\<Colon>linordered_idom. 1 * x\<^isub>1 = x\<^isub>1" by (metis mult_1)
    1.59 -  have F2: "\<forall>x\<^isub>3 x\<^isub>1\<Colon>'a\<Colon>linordered_idom. 0 \<le> x\<^isub>1 \<longrightarrow> \<bar>x\<^isub>3 * x\<^isub>1\<bar> = \<bar>x\<^isub>3\<bar> * x\<^isub>1" by (metis abs_mult_pos)
    1.60 -  hence "\<forall>x\<^isub>1\<ge>0. \<bar>x\<^isub>1\<Colon>'a\<Colon>linordered_idom\<bar> = x\<^isub>1" by (metis F1 abs_one)
    1.61 -  hence "\<forall>x\<^isub>3. 0 \<le> \<bar>f x\<^isub>3\<bar> \<longrightarrow> c * \<bar>f x\<^isub>3\<bar> = \<bar>c\<bar> * \<bar>f x\<^isub>3\<bar>" by (metis F2 A1 abs_ge_zero order_trans)
    1.62 +  have F1: "\<forall>x\<^sub>1\<Colon>'a\<Colon>linordered_idom. 1 * x\<^sub>1 = x\<^sub>1" by (metis mult_1)
    1.63 +  have F2: "\<forall>x\<^sub>3 x\<^sub>1\<Colon>'a\<Colon>linordered_idom. 0 \<le> x\<^sub>1 \<longrightarrow> \<bar>x\<^sub>3 * x\<^sub>1\<bar> = \<bar>x\<^sub>3\<bar> * x\<^sub>1" by (metis abs_mult_pos)
    1.64 +  hence "\<forall>x\<^sub>1\<ge>0. \<bar>x\<^sub>1\<Colon>'a\<Colon>linordered_idom\<bar> = x\<^sub>1" by (metis F1 abs_one)
    1.65 +  hence "\<forall>x\<^sub>3. 0 \<le> \<bar>f x\<^sub>3\<bar> \<longrightarrow> c * \<bar>f x\<^sub>3\<bar> = \<bar>c\<bar> * \<bar>f x\<^sub>3\<bar>" by (metis F2 A1 abs_ge_zero order_trans)
    1.66    thus "\<bar>h x\<bar> \<le> \<bar>c\<bar> * \<bar>f x\<bar>" by (metis A1 abs_ge_zero)
    1.67  qed
    1.68  
    1.69 @@ -116,8 +116,8 @@
    1.70  proof -
    1.71    fix c :: 'a and x :: 'b
    1.72    assume A1: "\<forall>x. \<bar>h x\<bar> \<le> c * \<bar>f x\<bar>"
    1.73 -  have "\<forall>x\<^isub>1\<Colon>'a\<Colon>linordered_idom. 1 * x\<^isub>1 = x\<^isub>1" by (metis mult_1)
    1.74 -  hence "\<forall>x\<^isub>3. \<bar>c * \<bar>f x\<^isub>3\<bar>\<bar> = c * \<bar>f x\<^isub>3\<bar>"
    1.75 +  have "\<forall>x\<^sub>1\<Colon>'a\<Colon>linordered_idom. 1 * x\<^sub>1 = x\<^sub>1" by (metis mult_1)
    1.76 +  hence "\<forall>x\<^sub>3. \<bar>c * \<bar>f x\<^sub>3\<bar>\<bar> = c * \<bar>f x\<^sub>3\<bar>"
    1.77      by (metis A1 abs_ge_zero order_trans abs_mult_pos abs_one)
    1.78    hence "\<bar>h x\<bar> \<le> \<bar>c * f x\<bar>" by (metis A1 abs_ge_zero abs_mult_pos abs_mult)
    1.79    thus "\<bar>h x\<bar> \<le> \<bar>c\<bar> * \<bar>f x\<bar>" by (metis abs_mult)
    1.80 @@ -459,11 +459,11 @@
    1.81    have "(0\<Colon>'a) < \<bar>c\<bar> \<longrightarrow> (0\<Colon>'a) < \<bar>inverse c\<bar>" using F1 by (metis positive_imp_inverse_positive)
    1.82    hence "(0\<Colon>'a) < \<bar>inverse c\<bar>" using F2 by metis
    1.83    hence F3: "(0\<Colon>'a) \<le> \<bar>inverse c\<bar>" by (metis order_le_less)
    1.84 -  have "\<exists>(u\<Colon>'a) SKF\<^isub>7\<Colon>'a \<Rightarrow> 'b. \<bar>g (SKF\<^isub>7 (\<bar>inverse c\<bar> * u))\<bar> \<le> u * \<bar>f (SKF\<^isub>7 (\<bar>inverse c\<bar> * u))\<bar>"
    1.85 +  have "\<exists>(u\<Colon>'a) SKF\<^sub>7\<Colon>'a \<Rightarrow> 'b. \<bar>g (SKF\<^sub>7 (\<bar>inverse c\<bar> * u))\<bar> \<le> u * \<bar>f (SKF\<^sub>7 (\<bar>inverse c\<bar> * u))\<bar>"
    1.86      using A2 by metis
    1.87 -  hence F4: "\<exists>(u\<Colon>'a) SKF\<^isub>7\<Colon>'a \<Rightarrow> 'b. \<bar>g (SKF\<^isub>7 (\<bar>inverse c\<bar> * u))\<bar> \<le> u * \<bar>f (SKF\<^isub>7 (\<bar>inverse c\<bar> * u))\<bar> \<and> (0\<Colon>'a) \<le> \<bar>inverse c\<bar>"
    1.88 +  hence F4: "\<exists>(u\<Colon>'a) SKF\<^sub>7\<Colon>'a \<Rightarrow> 'b. \<bar>g (SKF\<^sub>7 (\<bar>inverse c\<bar> * u))\<bar> \<le> u * \<bar>f (SKF\<^sub>7 (\<bar>inverse c\<bar> * u))\<bar> \<and> (0\<Colon>'a) \<le> \<bar>inverse c\<bar>"
    1.89      using F3 by metis
    1.90 -  hence "\<exists>(v\<Colon>'a) (u\<Colon>'a) SKF\<^isub>7\<Colon>'a \<Rightarrow> 'b. \<bar>inverse c\<bar> * \<bar>g (SKF\<^isub>7 (u * v))\<bar> \<le> u * (v * \<bar>f (SKF\<^isub>7 (u * v))\<bar>)"
    1.91 +  hence "\<exists>(v\<Colon>'a) (u\<Colon>'a) SKF\<^sub>7\<Colon>'a \<Rightarrow> 'b. \<bar>inverse c\<bar> * \<bar>g (SKF\<^sub>7 (u * v))\<bar> \<le> u * (v * \<bar>f (SKF\<^sub>7 (u * v))\<bar>)"
    1.92      by (metis comm_mult_left_mono)
    1.93    thus "\<exists>ca\<Colon>'a. \<forall>x\<Colon>'b. \<bar>inverse c\<bar> * \<bar>g x\<bar> \<le> ca * \<bar>f x\<bar>"
    1.94      using A2 F4 by (metis ab_semigroup_mult_class.mult_ac(1) comm_mult_left_mono)