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[] | lemma n_less_equal_power_2:
"n < 2 ^ n" by (fact less_exp) | lemma n_less_equal_power_2:
"n < 2 ^ n" by (fact less_exp) | proof (prove)
goal (1 subgoal):
1. n < 2 ^ n | lemma n_less_equal_power_2:
"n < 2 ^ n" | unnamed_thy_1 | More_Arithmetic | 1 |
|
[] | lemma min_pm [simp]: "min a b + (a - b) = a"
for a b :: nat by arith | lemma min_pm [simp]: "min a b + (a - b) = a"
for a b :: nat by arith | proof (prove)
goal (1 subgoal):
1. min a b + (a - b) = a | lemma min_pm [simp]: "min a b + (a - b) = a"
for a b :: nat | unnamed_thy_2 | More_Arithmetic | 1 |
|
[] | lemma min_pm1 [simp]: "a - b + min a b = a"
for a b :: nat by arith | lemma min_pm1 [simp]: "a - b + min a b = a"
for a b :: nat by arith | proof (prove)
goal (1 subgoal):
1. a - b + min a b = a | lemma min_pm1 [simp]: "a - b + min a b = a"
for a b :: nat | unnamed_thy_3 | More_Arithmetic | 1 |
|
[] | lemma rev_min_pm [simp]: "min b a + (a - b) = a"
for a b :: nat by arith | lemma rev_min_pm [simp]: "min b a + (a - b) = a"
for a b :: nat by arith | proof (prove)
goal (1 subgoal):
1. min b a + (a - b) = a | lemma rev_min_pm [simp]: "min b a + (a - b) = a"
for a b :: nat | unnamed_thy_4 | More_Arithmetic | 1 |
|
[] | lemma rev_min_pm1 [simp]: "a - b + min b a = a"
for a b :: nat by arith | lemma rev_min_pm1 [simp]: "a - b + min b a = a"
for a b :: nat by arith | proof (prove)
goal (1 subgoal):
1. a - b + min b a = a | lemma rev_min_pm1 [simp]: "a - b + min b a = a"
for a b :: nat | unnamed_thy_5 | More_Arithmetic | 1 |
|
[] | lemma min_minus' [simp]: "min (m - k) m = m - k"
for m k :: nat by arith | lemma min_minus' [simp]: "min (m - k) m = m - k"
for m k :: nat by arith | proof (prove)
goal (1 subgoal):
1. min (m - k) m = m - k | lemma min_minus' [simp]: "min (m - k) m = m - k"
for m k :: nat | unnamed_thy_7 | More_Arithmetic | 1 |
|
[] | lemma nat_less_power_trans:
fixes n :: nat
assumes nv: "n < 2 ^ (m - k)"
and kv: "k \<le> m"
shows "2 ^ k * n < 2 ^ m" proof (rule order_less_le_trans) show "2 ^ k * n < 2 ^ k * 2 ^ (m - k)" by (rule mult_less_mono2 [OF nv zero_less_power]) simp show "(2::nat) ^ k * 2 ^ (m - k) \<le> 2 ^ m" using nv kv by (subst power_add [symmetric]) simp qed | lemma nat_less_power_trans:
fixes n :: nat
assumes nv: "n < 2 ^ (m - k)"
and kv: "k \<le> m"
shows "2 ^ k * n < 2 ^ m" proof (rule order_less_le_trans) show "2 ^ k * n < 2 ^ k * 2 ^ (m - k)" by (rule mult_less_mono2 [OF nv zero_less_power]) simp show "(2::nat) ^ k * 2 ^ (m - k) \<le> 2 ^ m" using nv kv by (subst power_add [symmetric]) simp qed | proof (prove)
goal (1 subgoal):
1. 2 ^ k * n < 2 ^ m proof (state)
goal (2 subgoals):
1. 2 ^ k * n < ?y
2. ?y \<le> 2 ^ m proof (prove)
goal (1 subgoal):
1. 2 ^ k * n < 2 ^ k * 2 ^ (m - k) proof (state)
this:
2 ^ k * n < 2 ^ k * 2 ^ (m - k)
goal (1 subgoal):
1. 2 ^ k * 2 ^ (m - k) \<le> 2 ^ m proof (prove)
goal (1 subgoal):
1. 2 ^ k * 2 ^ (m - k) \<le> 2 ^ m proof (prove)
using this:
n < 2 ^ (m - k)
k \<le> m
goal (1 subgoal):
1. 2 ^ k * 2 ^ (m - k) \<le> 2 ^ m proof (state)
this:
2 ^ k * 2 ^ (m - k) \<le> 2 ^ m
goal:
No subgoals! | lemma nat_less_power_trans:
fixes n :: nat
assumes nv: "n < 2 ^ (m - k)"
and kv: "k \<le> m"
shows "2 ^ k * n < 2 ^ m" | unnamed_thy_8 | More_Arithmetic | 7 |
|
[] | lemma nat_le_power_trans:
fixes n :: nat
shows "\<lbrakk>n \<le> 2 ^ (m - k); k \<le> m\<rbrakk> \<Longrightarrow> 2 ^ k * n \<le> 2 ^ m" by (metis le_add_diff_inverse mult_le_mono2 semiring_normalization_rules(26)) | lemma nat_le_power_trans:
fixes n :: nat
shows "\<lbrakk>n \<le> 2 ^ (m - k); k \<le> m\<rbrakk> \<Longrightarrow> 2 ^ k * n \<le> 2 ^ m" by (metis le_add_diff_inverse mult_le_mono2 semiring_normalization_rules(26)) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>n \<le> 2 ^ (m - k); k \<le> m\<rbrakk> \<Longrightarrow> 2 ^ k * n \<le> 2 ^ m | lemma nat_le_power_trans:
fixes n :: nat
shows "\<lbrakk>n \<le> 2 ^ (m - k); k \<le> m\<rbrakk> \<Longrightarrow> 2 ^ k * n \<le> 2 ^ m" | unnamed_thy_9 | More_Arithmetic | 1 |
|
[] | lemma nat_add_offset_less:
fixes x :: nat
assumes yv: "y < 2 ^ n"
and xv: "x < 2 ^ m"
and mn: "sz = m + n"
shows "x * 2 ^ n + y < 2 ^ sz" proof (subst mn) from yv obtain qy where "y + qy = 2 ^ n" and "0 < qy" by (auto dest: less_imp_add_positive) have "x * 2 ^ n + y < x * 2 ^ n + 2 ^ n" by simp fact+ also have "\<dots> = (x + 1) * 2 ^ n" by simp also have "\<dots> \<le> 2 ^ (m + n)" using xv by (subst power_add) (rule mult_le_mono1, simp) finally show "x * 2 ^ n + y < 2 ^ (m + n)" . qed | lemma nat_add_offset_less:
fixes x :: nat
assumes yv: "y < 2 ^ n"
and xv: "x < 2 ^ m"
and mn: "sz = m + n"
shows "x * 2 ^ n + y < 2 ^ sz" proof (subst mn) from yv obtain qy where "y + qy = 2 ^ n" and "0 < qy" by (auto dest: less_imp_add_positive) have "x * 2 ^ n + y < x * 2 ^ n + 2 ^ n" by simp fact+ also have "\<dots> = (x + 1) * 2 ^ n" by simp also have "\<dots> \<le> 2 ^ (m + n)" using xv by (subst power_add) (rule mult_le_mono1, simp) finally show "x * 2 ^ n + y < 2 ^ (m + n)" . qed | proof (prove)
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ sz proof (state)
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ (m + n) proof (chain)
picking this:
y < 2 ^ n proof (prove)
using this:
y < 2 ^ n
goal (1 subgoal):
1. (\<And>qy. \<lbrakk>y + qy = 2 ^ n; 0 < qy\<rbrakk> \<Longrightarrow> thesis) \<Longrightarrow> thesis proof (state)
this:
y + qy = 2 ^ n
0 < qy
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ (m + n) proof (prove)
goal (1 subgoal):
1. x * 2 ^ n + y < x * 2 ^ n + 2 ^ n proof (state)
this:
x * 2 ^ n + y < x * 2 ^ n + 2 ^ n
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ (m + n) proof (state)
this:
x * 2 ^ n + y < x * 2 ^ n + 2 ^ n
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ (m + n) proof (prove)
goal (1 subgoal):
1. x * 2 ^ n + 2 ^ n = (x + 1) * 2 ^ n proof (state)
this:
x * 2 ^ n + 2 ^ n = (x + 1) * 2 ^ n
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ (m + n) proof (state)
this:
x * 2 ^ n + 2 ^ n = (x + 1) * 2 ^ n
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ (m + n) proof (prove)
goal (1 subgoal):
1. (x + 1) * 2 ^ n \<le> 2 ^ (m + n) proof (prove)
using this:
x < 2 ^ m
goal (1 subgoal):
1. (x + 1) * 2 ^ n \<le> 2 ^ (m + n) proof (state)
this:
(x + 1) * 2 ^ n \<le> 2 ^ (m + n)
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ (m + n) proof (chain)
picking this:
x * 2 ^ n + y < 2 ^ (m + n) proof (prove)
using this:
x * 2 ^ n + y < 2 ^ (m + n)
goal (1 subgoal):
1. x * 2 ^ n + y < 2 ^ (m + n) proof (state)
this:
x * 2 ^ n + y < 2 ^ (m + n)
goal:
No subgoals! | lemma nat_add_offset_less:
fixes x :: nat
assumes yv: "y < 2 ^ n"
and xv: "x < 2 ^ m"
and mn: "sz = m + n"
shows "x * 2 ^ n + y < 2 ^ sz" | unnamed_thy_10 | More_Arithmetic | 17 |
|
[] | lemma nat_power_less_diff:
assumes lt: "(2::nat) ^ n * q < 2 ^ m"
shows "q < 2 ^ (m - n)" using lt proof (induct n arbitrary: m) case 0 then show ?case by simp next case (Suc n) have ih: "\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n)"
and prem: "2 ^ Suc n * q < 2 ^ m" by fact+ show ?case proof (cases m) case 0 then show ?thesis using Suc by simp next case (Suc m') then show ?thesis using prem by (simp add: ac_simps ih) qed qed | lemma nat_power_less_diff:
assumes lt: "(2::nat) ^ n * q < 2 ^ m"
shows "q < 2 ^ (m - n)" using lt proof (induct n arbitrary: m) case 0 then show ?case by simp next case (Suc n) have ih: "\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n)"
and prem: "2 ^ Suc n * q < 2 ^ m" by fact+ show ?case proof (cases m) case 0 then show ?thesis using Suc by simp next case (Suc m') then show ?thesis using prem by (simp add: ac_simps ih) qed qed | proof (prove)
goal (1 subgoal):
1. q < 2 ^ (m - n) proof (prove)
using this:
2 ^ n * q < 2 ^ m
goal (1 subgoal):
1. q < 2 ^ (m - n) proof (state)
goal (2 subgoals):
1. \<And>m. 2 ^ 0 * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - 0)
2. \<And>n m. \<lbrakk>\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n); 2 ^ Suc n * q < 2 ^ m\<rbrakk> \<Longrightarrow> q < 2 ^ (m - Suc n) proof (state)
this:
2 ^ 0 * q < 2 ^ m
goal (2 subgoals):
1. \<And>m. 2 ^ 0 * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - 0)
2. \<And>n m. \<lbrakk>\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n); 2 ^ Suc n * q < 2 ^ m\<rbrakk> \<Longrightarrow> q < 2 ^ (m - Suc n) proof (chain)
picking this:
2 ^ 0 * q < 2 ^ m proof (prove)
using this:
2 ^ 0 * q < 2 ^ m
goal (1 subgoal):
1. q < 2 ^ (m - 0) proof (state)
this:
q < 2 ^ (m - 0)
goal (1 subgoal):
1. \<And>n m. \<lbrakk>\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n); 2 ^ Suc n * q < 2 ^ m\<rbrakk> \<Longrightarrow> q < 2 ^ (m - Suc n) proof (state)
goal (1 subgoal):
1. \<And>n m. \<lbrakk>\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n); 2 ^ Suc n * q < 2 ^ m\<rbrakk> \<Longrightarrow> q < 2 ^ (m - Suc n) proof (state)
this:
2 ^ n * q < 2 ^ ?m \<Longrightarrow> q < 2 ^ (?m - n)
2 ^ Suc n * q < 2 ^ m
goal (1 subgoal):
1. \<And>n m. \<lbrakk>\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n); 2 ^ Suc n * q < 2 ^ m\<rbrakk> \<Longrightarrow> q < 2 ^ (m - Suc n) proof (prove)
goal (1 subgoal):
1. (\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n)) &&& 2 ^ Suc n * q < 2 ^ m proof (state)
this:
2 ^ n * q < 2 ^ ?m \<Longrightarrow> q < 2 ^ (?m - n)
2 ^ Suc n * q < 2 ^ m
goal (1 subgoal):
1. \<And>n m. \<lbrakk>\<And>m. 2 ^ n * q < 2 ^ m \<Longrightarrow> q < 2 ^ (m - n); 2 ^ Suc n * q < 2 ^ m\<rbrakk> \<Longrightarrow> q < 2 ^ (m - Suc n) proof (prove)
goal (1 subgoal):
1. q < 2 ^ (m - Suc n) proof (state)
goal (2 subgoals):
1. m = 0 \<Longrightarrow> q < 2 ^ (m - Suc n)
2. \<And>nat. m = Suc nat \<Longrightarrow> q < 2 ^ (m - Suc n) proof (state)
this:
m = 0
goal (2 subgoals):
1. m = 0 \<Longrightarrow> q < 2 ^ (m - Suc n)
2. \<And>nat. m = Suc nat \<Longrightarrow> q < 2 ^ (m - Suc n) proof (chain)
picking this:
m = 0 proof (prove)
using this:
m = 0
goal (1 subgoal):
1. q < 2 ^ (m - Suc n) proof (prove)
using this:
m = 0
2 ^ n * q < 2 ^ ?m \<Longrightarrow> q < 2 ^ (?m - n)
2 ^ Suc n * q < 2 ^ m
goal (1 subgoal):
1. q < 2 ^ (m - Suc n) proof (state)
this:
q < 2 ^ (m - Suc n)
goal (1 subgoal):
1. \<And>nat. m = Suc nat \<Longrightarrow> q < 2 ^ (m - Suc n) proof (state)
goal (1 subgoal):
1. \<And>nat. m = Suc nat \<Longrightarrow> q < 2 ^ (m - Suc n) proof (state)
this:
m = Suc m'
goal (1 subgoal):
1. \<And>nat. m = Suc nat \<Longrightarrow> q < 2 ^ (m - Suc n) proof (chain)
picking this:
m = Suc m' proof (prove)
using this:
m = Suc m'
goal (1 subgoal):
1. q < 2 ^ (m - Suc n) proof (prove)
using this:
m = Suc m'
2 ^ Suc n * q < 2 ^ m
goal (1 subgoal):
1. q < 2 ^ (m - Suc n) proof (state)
this:
q < 2 ^ (m - Suc n)
goal:
No subgoals! proof (state)
this:
q < 2 ^ (m - Suc n)
goal:
No subgoals! | lemma nat_power_less_diff:
assumes lt: "(2::nat) ^ n * q < 2 ^ m"
shows "q < 2 ^ (m - n)" | unnamed_thy_11 | More_Arithmetic | 25 |
|
[] | lemma power_2_mult_step_le:
"\<lbrakk>n' \<le> n; 2 ^ n' * k' < 2 ^ n * k\<rbrakk> \<Longrightarrow> 2 ^ n' * (k' + 1) \<le> 2 ^ n * (k::nat)" apply (cases "n'=n", simp) apply (metis Suc_leI le_refl mult_Suc_right mult_le_mono semiring_normalization_rules(7)) apply (drule (1) le_neq_trans) apply clarsimp apply (subgoal_tac "\<exists>m. n = n' + m") prefer 2 apply (simp add: le_Suc_ex) apply (clarsimp simp: power_add) apply (metis Suc_leI mult.assoc mult_Suc_right nat_mult_le_cancel_disj) done | lemma power_2_mult_step_le:
"\<lbrakk>n' \<le> n; 2 ^ n' * k' < 2 ^ n * k\<rbrakk> \<Longrightarrow> 2 ^ n' * (k' + 1) \<le> 2 ^ n * (k::nat)" apply (cases "n'=n", simp) apply (metis Suc_leI le_refl mult_Suc_right mult_le_mono semiring_normalization_rules(7)) apply (drule (1) le_neq_trans) apply clarsimp apply (subgoal_tac "\<exists>m. n = n' + m") prefer 2 apply (simp add: le_Suc_ex) apply (clarsimp simp: power_add) apply (metis Suc_leI mult.assoc mult_Suc_right nat_mult_le_cancel_disj) done | proof (prove)
goal (1 subgoal):
1. \<lbrakk>n' \<le> n; 2 ^ n' * k' < 2 ^ n * k\<rbrakk> \<Longrightarrow> 2 ^ n' * (k' + 1) \<le> 2 ^ n * k proof (prove)
goal (2 subgoals):
1. \<lbrakk>k' < k; n' = n\<rbrakk> \<Longrightarrow> 2 ^ n + 2 ^ n * k' \<le> 2 ^ n * k
2. \<lbrakk>n' \<le> n; 2 ^ n' * k' < 2 ^ n * k; n' \<noteq> n\<rbrakk> \<Longrightarrow> 2 ^ n' * (k' + 1) \<le> 2 ^ n * k proof (prove)
goal (1 subgoal):
1. \<lbrakk>n' \<le> n; 2 ^ n' * k' < 2 ^ n * k; n' \<noteq> n\<rbrakk> \<Longrightarrow> 2 ^ n' * (k' + 1) \<le> 2 ^ n * k proof (prove)
goal (1 subgoal):
1. \<lbrakk>2 ^ n' * k' < 2 ^ n * k; n' \<noteq> n; n' < n\<rbrakk> \<Longrightarrow> 2 ^ n' * (k' + 1) \<le> 2 ^ n * k proof (prove)
goal (1 subgoal):
1. \<lbrakk>2 ^ n' * k' < 2 ^ n * k; n' < n\<rbrakk> \<Longrightarrow> 2 ^ n' + 2 ^ n' * k' \<le> 2 ^ n * k proof (prove)
goal (2 subgoals):
1. \<lbrakk>2 ^ n' * k' < 2 ^ n * k; n' < n; \<exists>m. n = n' + m\<rbrakk> \<Longrightarrow> 2 ^ n' + 2 ^ n' * k' \<le> 2 ^ n * k
2. \<lbrakk>2 ^ n' * k' < 2 ^ n * k; n' < n\<rbrakk> \<Longrightarrow> \<exists>m. n = n' + m proof (prove)
goal (2 subgoals):
1. \<lbrakk>2 ^ n' * k' < 2 ^ n * k; n' < n\<rbrakk> \<Longrightarrow> \<exists>m. n = n' + m
2. \<lbrakk>2 ^ n' * k' < 2 ^ n * k; n' < n; \<exists>m. n = n' + m\<rbrakk> \<Longrightarrow> 2 ^ n' + 2 ^ n' * k' \<le> 2 ^ n * k proof (prove)
goal (1 subgoal):
1. \<lbrakk>2 ^ n' * k' < 2 ^ n * k; n' < n; \<exists>m. n = n' + m\<rbrakk> \<Longrightarrow> 2 ^ n' + 2 ^ n' * k' \<le> 2 ^ n * k proof (prove)
goal (1 subgoal):
1. \<And>m. \<lbrakk>k' < 2 ^ m * k; 0 < m; n = n' + m\<rbrakk> \<Longrightarrow> 2 ^ n' + 2 ^ n' * k' \<le> 2 ^ n' * 2 ^ m * k proof (prove)
goal:
No subgoals! | lemma power_2_mult_step_le:
"\<lbrakk>n' \<le> n; 2 ^ n' * k' < 2 ^ n * k\<rbrakk> \<Longrightarrow> 2 ^ n' * (k' + 1) \<le> 2 ^ n * (k::nat)" | unnamed_thy_12 | More_Arithmetic | 10 |
|
[] | lemma nat_mult_power_less_eq:
"b > 0 \<Longrightarrow> (a * b ^ n < (b :: nat) ^ m) = (a < b ^ (m - n))" using mult_less_cancel2[where m = a and k = "b ^ n" and n="b ^ (m - n)"]
mult_less_cancel2[where m="a * b ^ (n - m)" and k="b ^ m" and n=1] apply (simp only: power_add[symmetric] nat_minus_add_max) apply (simp only: power_add[symmetric] nat_minus_add_max ac_simps) apply (simp add: max_def split: if_split_asm) done | lemma nat_mult_power_less_eq:
"b > 0 \<Longrightarrow> (a * b ^ n < (b :: nat) ^ m) = (a < b ^ (m - n))" using mult_less_cancel2[where m = a and k = "b ^ n" and n="b ^ (m - n)"]
mult_less_cancel2[where m="a * b ^ (n - m)" and k="b ^ m" and n=1] apply (simp only: power_add[symmetric] nat_minus_add_max) apply (simp only: power_add[symmetric] nat_minus_add_max ac_simps) apply (simp add: max_def split: if_split_asm) done | proof (prove)
goal (1 subgoal):
1. 0 < b \<Longrightarrow> (a * b ^ n < b ^ m) = (a < b ^ (m - n)) proof (prove)
using this:
(a * b ^ n < b ^ (m - n) * b ^ n) = (0 < b ^ n \<and> a < b ^ (m - n))
(a * b ^ (n - m) * b ^ m < 1 * b ^ m) = (0 < b ^ m \<and> a * b ^ (n - m) < 1)
goal (1 subgoal):
1. 0 < b \<Longrightarrow> (a * b ^ n < b ^ m) = (a < b ^ (m - n)) proof (prove)
goal (1 subgoal):
1. \<lbrakk>0 < b; (a * b ^ n < b ^ max m n) = (0 < b ^ n \<and> a < b ^ (m - n)); (a * b ^ (n - m) * b ^ m < 1 * b ^ m) = (0 < b ^ m \<and> a * b ^ (n - m) < 1)\<rbrakk> \<Longrightarrow> (a * b ^ n < b ^ m) = (a < b ^ (m - n)) proof (prove)
goal (1 subgoal):
1. \<lbrakk>0 < b; (a * b ^ n < b ^ max m n) = (0 < b ^ n \<and> a < b ^ (m - n)); (a * b ^ (m + (n - m)) < 1 * b ^ m) = (0 < b ^ m \<and> a * b ^ (n - m) < 1)\<rbrakk> \<Longrightarrow> (a * b ^ n < b ^ m) = (a < b ^ (m - n)) proof (prove)
goal:
No subgoals! | lemma nat_mult_power_less_eq:
"b > 0 \<Longrightarrow> (a * b ^ n < (b :: nat) ^ m) = (a < b ^ (m - n))" | unnamed_thy_13 | More_Arithmetic | 5 |
|
[] | lemma diff_diff_less:
"(i < m - (m - (n :: nat))) = (i < m \<and> i < n)" by auto | lemma diff_diff_less:
"(i < m - (m - (n :: nat))) = (i < m \<and> i < n)" by auto | proof (prove)
goal (1 subgoal):
1. (i < m - (m - n)) = (i < m \<and> i < n) | lemma diff_diff_less:
"(i < m - (m - (n :: nat))) = (i < m \<and> i < n)" | unnamed_thy_14 | More_Arithmetic | 1 |
|
[] | lemma small_powers_of_2:
\<open>x < 2 ^ (x - 1)\<close> if \<open>x \<ge> 3\<close> for x :: nat proof - define m where \<open>m = x - 3\<close> with that have \<open>x = m + 3\<close> by simp moreover have \<open>m + 3 < 4 * 2 ^ m\<close> by (induction m) simp_all ultimately show ?thesis by simp qed | lemma small_powers_of_2:
\<open>x < 2 ^ (x - 1)\<close> if \<open>x \<ge> 3\<close> for x :: nat proof - define m where \<open>m = x - 3\<close> with that have \<open>x = m + 3\<close> by simp moreover have \<open>m + 3 < 4 * 2 ^ m\<close> by (induction m) simp_all ultimately show ?thesis by simp qed | proof (prove)
goal (1 subgoal):
1. x < 2 ^ (x - 1) proof (state)
goal (1 subgoal):
1. x < 2 ^ (x - 1) proof (state)
this:
m = x - 3
goal (1 subgoal):
1. x < 2 ^ (x - 1) proof (chain)
picking this:
3 \<le> x
m = x - 3 proof (prove)
using this:
3 \<le> x
m = x - 3
goal (1 subgoal):
1. x = m + 3 proof (state)
this:
x = m + 3
goal (1 subgoal):
1. x < 2 ^ (x - 1) proof (state)
this:
x = m + 3
goal (1 subgoal):
1. x < 2 ^ (x - 1) proof (prove)
goal (1 subgoal):
1. m + 3 < 4 * 2 ^ m proof (state)
this:
m + 3 < 4 * 2 ^ m
goal (1 subgoal):
1. x < 2 ^ (x - 1) proof (chain)
picking this:
x = m + 3
m + 3 < 4 * 2 ^ m proof (prove)
using this:
x = m + 3
m + 3 < 4 * 2 ^ m
goal (1 subgoal):
1. x < 2 ^ (x - 1) proof (state)
this:
x < 2 ^ (x - 1)
goal:
No subgoals! | lemma small_powers_of_2:
\<open>x < 2 ^ (x - 1)\<close> if \<open>x \<ge> 3\<close> for x :: nat | unnamed_thy_15 | More_Arithmetic | 12 |
|
[] | lemma msrevs:
"0 < n \<Longrightarrow> (k * n + m) div n = m div n + k"
"(k * n + m) mod n = m mod n"
for n :: nat by simp_all | lemma msrevs:
"0 < n \<Longrightarrow> (k * n + m) div n = m div n + k"
"(k * n + m) mod n = m mod n"
for n :: nat by simp_all | proof (prove)
goal (1 subgoal):
1. (0 < n \<Longrightarrow> (k * n + m) div n = m div n + k) &&& (k * n + m) mod n = m mod n | lemma msrevs:
"0 < n \<Longrightarrow> (k * n + m) div n = m div n + k"
"(k * n + m) mod n = m mod n"
for n :: nat | unnamed_thy_16 | More_Arithmetic | 1 |
|
[] | lemma int_div_same_is_1 [simp]:
\<open>a div b = a \<longleftrightarrow> b = 1\<close> if \<open>0 < a\<close> for a b :: int using that by (metis div_by_1 abs_ge_zero abs_of_pos int_div_less_self neq_iff
nonneg1_imp_zdiv_pos_iff zabs_less_one_iff) | lemma int_div_same_is_1 [simp]:
\<open>a div b = a \<longleftrightarrow> b = 1\<close> if \<open>0 < a\<close> for a b :: int using that by (metis div_by_1 abs_ge_zero abs_of_pos int_div_less_self neq_iff
nonneg1_imp_zdiv_pos_iff zabs_less_one_iff) | proof (prove)
goal (1 subgoal):
1. (a div b = a) = (b = 1) proof (prove)
using this:
0 < a
goal (1 subgoal):
1. (a div b = a) = (b = 1) | lemma int_div_same_is_1 [simp]:
\<open>a div b = a \<longleftrightarrow> b = 1\<close> if \<open>0 < a\<close> for a b :: int | unnamed_thy_17 | More_Divides | 2 |
|
[] | lemma int_div_minus_is_minus1 [simp]:
\<open>a div b = - a \<longleftrightarrow> b = - 1\<close> if \<open>0 > a\<close> for a b :: int using that by (metis div_minus_right equation_minus_iff int_div_same_is_1 neg_0_less_iff_less) | lemma int_div_minus_is_minus1 [simp]:
\<open>a div b = - a \<longleftrightarrow> b = - 1\<close> if \<open>0 > a\<close> for a b :: int using that by (metis div_minus_right equation_minus_iff int_div_same_is_1 neg_0_less_iff_less) | proof (prove)
goal (1 subgoal):
1. (a div b = - a) = (b = - 1) proof (prove)
using this:
a < 0
goal (1 subgoal):
1. (a div b = - a) = (b = - 1) | lemma int_div_minus_is_minus1 [simp]:
\<open>a div b = - a \<longleftrightarrow> b = - 1\<close> if \<open>0 > a\<close> for a b :: int | unnamed_thy_18 | More_Divides | 2 |
|
[] | lemma nat_div_eq_Suc_0_iff: "n div m = Suc 0 \<longleftrightarrow> m \<le> n \<and> n < 2 * m" apply auto using div_greater_zero_iff apply fastforce apply (metis One_nat_def div_greater_zero_iff dividend_less_div_times mult.right_neutral mult_Suc mult_numeral_1 numeral_2_eq_2 zero_less_numeral) apply (simp add: div_nat_eqI) done | lemma nat_div_eq_Suc_0_iff: "n div m = Suc 0 \<longleftrightarrow> m \<le> n \<and> n < 2 * m" apply auto using div_greater_zero_iff apply fastforce apply (metis One_nat_def div_greater_zero_iff dividend_less_div_times mult.right_neutral mult_Suc mult_numeral_1 numeral_2_eq_2 zero_less_numeral) apply (simp add: div_nat_eqI) done | proof (prove)
goal (1 subgoal):
1. (n div m = Suc 0) = (m \<le> n \<and> n < 2 * m) proof (prove)
goal (3 subgoals):
1. n div m = Suc 0 \<Longrightarrow> m \<le> n
2. n div m = Suc 0 \<Longrightarrow> n < 2 * m
3. \<lbrakk>m \<le> n; n < 2 * m\<rbrakk> \<Longrightarrow> n div m = Suc 0 proof (prove)
using this:
(0 < ?m div ?n) = (?n \<le> ?m \<and> 0 < ?n)
goal (3 subgoals):
1. n div m = Suc 0 \<Longrightarrow> m \<le> n
2. n div m = Suc 0 \<Longrightarrow> n < 2 * m
3. \<lbrakk>m \<le> n; n < 2 * m\<rbrakk> \<Longrightarrow> n div m = Suc 0 proof (prove)
goal (2 subgoals):
1. n div m = Suc 0 \<Longrightarrow> n < 2 * m
2. \<lbrakk>m \<le> n; n < 2 * m\<rbrakk> \<Longrightarrow> n div m = Suc 0 proof (prove)
goal (1 subgoal):
1. \<lbrakk>m \<le> n; n < 2 * m\<rbrakk> \<Longrightarrow> n div m = Suc 0 proof (prove)
goal:
No subgoals! | lemma nat_div_eq_Suc_0_iff: "n div m = Suc 0 \<longleftrightarrow> m \<le> n \<and> n < 2 * m" | unnamed_thy_19 | More_Divides | 6 |
|
[] | lemma diff_mod_le:
\<open>a - a mod b \<le> d - b\<close> if \<open>a < d\<close> \<open>b dvd d\<close> for a b d :: nat using that apply(subst minus_mod_eq_mult_div) apply(clarsimp simp: dvd_def) apply(cases \<open>b = 0\<close>) apply simp apply(subgoal_tac "a div b \<le> k - 1") prefer 2 apply(subgoal_tac "a div b < k") apply(simp add: less_Suc_eq_le [symmetric]) apply(subgoal_tac "b * (a div b) < b * ((b * k) div b)") apply clarsimp apply(subst div_mult_self1_is_m) apply arith apply(rule le_less_trans) apply simp apply(subst mult.commute) apply(rule div_times_less_eq_dividend) apply assumption apply clarsimp apply(subgoal_tac "b * (a div b) \<le> b * (k - 1)") apply(erule le_trans) apply(simp add: diff_mult_distrib2) apply simp done | lemma diff_mod_le:
\<open>a - a mod b \<le> d - b\<close> if \<open>a < d\<close> \<open>b dvd d\<close> for a b d :: nat using that apply(subst minus_mod_eq_mult_div) apply(clarsimp simp: dvd_def) apply(cases \<open>b = 0\<close>) apply simp apply(subgoal_tac "a div b \<le> k - 1") prefer 2 apply(subgoal_tac "a div b < k") apply(simp add: less_Suc_eq_le [symmetric]) apply(subgoal_tac "b * (a div b) < b * ((b * k) div b)") apply clarsimp apply(subst div_mult_self1_is_m) apply arith apply(rule le_less_trans) apply simp apply(subst mult.commute) apply(rule div_times_less_eq_dividend) apply assumption apply clarsimp apply(subgoal_tac "b * (a div b) \<le> b * (k - 1)") apply(erule le_trans) apply(simp add: diff_mult_distrib2) apply simp done | proof (prove)
goal (1 subgoal):
1. a - a mod b \<le> d - b proof (prove)
using this:
a < d
b dvd d
goal (1 subgoal):
1. a - a mod b \<le> d - b proof (prove)
goal (1 subgoal):
1. \<lbrakk>a < d; b dvd d\<rbrakk> \<Longrightarrow> b * (a div b) \<le> d - b proof (prove)
goal (1 subgoal):
1. \<And>k. \<lbrakk>a < b * k; d = b * k\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b = 0\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (1 subgoal):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> a div b \<le> k - 1 proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> a div b \<le> k - 1
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (3 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b < k\<rbrakk> \<Longrightarrow> a div b \<le> k - 1
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> a div b < k
3. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> a div b < k
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (3 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; b * (a div b) < b * (b * k div b)\<rbrakk> \<Longrightarrow> a div b < k
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> b * (a div b) < b * (b * k div b)
3. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> b * (a div b) < b * (b * k div b)
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (3 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> 0 < b
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> b * (a div b) < b * k
3. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> b * (a div b) < b * k
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (3 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> b * (a div b) \<le> ?y17 k
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> ?y17 k < b * k
3. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (3 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; 0 < b\<rbrakk> \<Longrightarrow> b * (a div b) \<le> ?y17 k
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> ?y17 k < b * k
3. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (3 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; 0 < b\<rbrakk> \<Longrightarrow> a div b * b \<le> ?y21 k (?k21 k)
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> ?y21 k (?k21 k) < b * k
3. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0\<rbrakk> \<Longrightarrow> a < b * k
2. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (1 subgoal):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; b \<noteq> 0; a div b \<le> k - 1\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (1 subgoal):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; 0 < b; a div b \<le> k - Suc 0\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; 0 < b; a div b \<le> k - Suc 0; b * (a div b) \<le> b * (k - 1)\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * k - b
2. \<And>k. \<lbrakk>a < b * k; d = b * k; 0 < b; a div b \<le> k - Suc 0\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * (k - 1) proof (prove)
goal (2 subgoals):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; 0 < b; a div b \<le> k - Suc 0\<rbrakk> \<Longrightarrow> b * (k - 1) \<le> b * k - b
2. \<And>k. \<lbrakk>a < b * k; d = b * k; 0 < b; a div b \<le> k - Suc 0\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * (k - 1) proof (prove)
goal (1 subgoal):
1. \<And>k. \<lbrakk>a < b * k; d = b * k; 0 < b; a div b \<le> k - Suc 0\<rbrakk> \<Longrightarrow> b * (a div b) \<le> b * (k - 1) proof (prove)
goal:
No subgoals! | lemma diff_mod_le:
\<open>a - a mod b \<le> d - b\<close> if \<open>a < d\<close> \<open>b dvd d\<close> for a b d :: nat | unnamed_thy_20 | More_Divides | 24 |
|
[] | lemma one_mod_exp_eq_one [simp]:
"1 mod (2 * 2 ^ n) = (1::int)" using power_gt1 [of 2 n] by (auto intro: mod_pos_pos_trivial) | lemma one_mod_exp_eq_one [simp]:
"1 mod (2 * 2 ^ n) = (1::int)" using power_gt1 [of 2 n] by (auto intro: mod_pos_pos_trivial) | proof (prove)
goal (1 subgoal):
1. 1 mod (2 * 2 ^ n) = 1 proof (prove)
using this:
(1::?'a1) < (2::?'a1) \<Longrightarrow> (1::?'a1) < (2::?'a1) ^ Suc n
goal (1 subgoal):
1. 1 mod (2 * 2 ^ n) = 1 | lemma one_mod_exp_eq_one [simp]:
"1 mod (2 * 2 ^ n) = (1::int)" | unnamed_thy_21 | More_Divides | 2 |
|
[] | lemma int_mod_lem: "0 < n \<Longrightarrow> 0 \<le> b \<and> b < n \<longleftrightarrow> b mod n = b"
for b n :: int apply safe apply (erule (1) mod_pos_pos_trivial) apply (erule_tac [!] subst) apply auto done | lemma int_mod_lem: "0 < n \<Longrightarrow> 0 \<le> b \<and> b < n \<longleftrightarrow> b mod n = b"
for b n :: int apply safe apply (erule (1) mod_pos_pos_trivial) apply (erule_tac [!] subst) apply auto done | proof (prove)
goal (1 subgoal):
1. 0 < n \<Longrightarrow> (0 \<le> b \<and> b < n) = (b mod n = b) proof (prove)
goal (3 subgoals):
1. \<lbrakk>0 < n; 0 \<le> b; b < n\<rbrakk> \<Longrightarrow> b mod n = b
2. \<lbrakk>0 < n; b mod n = b\<rbrakk> \<Longrightarrow> 0 \<le> b
3. \<lbrakk>0 < n; b mod n = b\<rbrakk> \<Longrightarrow> b < n proof (prove)
goal (2 subgoals):
1. \<lbrakk>0 < n; b mod n = b\<rbrakk> \<Longrightarrow> 0 \<le> b
2. \<lbrakk>0 < n; b mod n = b\<rbrakk> \<Longrightarrow> b < n proof (prove)
goal (2 subgoals):
1. 0 < n \<Longrightarrow> 0 \<le> b mod n
2. 0 < n \<Longrightarrow> b mod n < n proof (prove)
goal:
No subgoals! | lemma int_mod_lem: "0 < n \<Longrightarrow> 0 \<le> b \<and> b < n \<longleftrightarrow> b mod n = b"
for b n :: int | unnamed_thy_22 | More_Divides | 5 |
|
[] | lemma int_mod_ge': "b < 0 \<Longrightarrow> 0 < n \<Longrightarrow> b + n \<le> b mod n"
for b n :: int by (metis add_less_same_cancel2 int_mod_ge mod_add_self2) | lemma int_mod_ge': "b < 0 \<Longrightarrow> 0 < n \<Longrightarrow> b + n \<le> b mod n"
for b n :: int by (metis add_less_same_cancel2 int_mod_ge mod_add_self2) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>b < 0; 0 < n\<rbrakk> \<Longrightarrow> b + n \<le> b mod n | lemma int_mod_ge': "b < 0 \<Longrightarrow> 0 < n \<Longrightarrow> b + n \<le> b mod n"
for b n :: int | unnamed_thy_23 | More_Divides | 1 |
|
[] | lemma int_mod_le': "0 \<le> b - n \<Longrightarrow> b mod n \<le> b - n"
for b n :: int by (metis minus_mod_self2 zmod_le_nonneg_dividend) | lemma int_mod_le': "0 \<le> b - n \<Longrightarrow> b mod n \<le> b - n"
for b n :: int by (metis minus_mod_self2 zmod_le_nonneg_dividend) | proof (prove)
goal (1 subgoal):
1. 0 \<le> b - n \<Longrightarrow> b mod n \<le> b - n | lemma int_mod_le': "0 \<le> b - n \<Longrightarrow> b mod n \<le> b - n"
for b n :: int | unnamed_thy_24 | More_Divides | 1 |
|
[] | lemma emep1: "even n \<Longrightarrow> even d \<Longrightarrow> 0 \<le> d \<Longrightarrow> (n + 1) mod d = (n mod d) + 1"
for n d :: int by (auto simp add: pos_zmod_mult_2 add.commute dvd_def) | lemma emep1: "even n \<Longrightarrow> even d \<Longrightarrow> 0 \<le> d \<Longrightarrow> (n + 1) mod d = (n mod d) + 1"
for n d :: int by (auto simp add: pos_zmod_mult_2 add.commute dvd_def) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>even n; even d; 0 \<le> d\<rbrakk> \<Longrightarrow> (n + 1) mod d = n mod d + 1 | lemma emep1: "even n \<Longrightarrow> even d \<Longrightarrow> 0 \<le> d \<Longrightarrow> (n + 1) mod d = (n mod d) + 1"
for n d :: int | unnamed_thy_25 | More_Divides | 1 |
|
[] | lemma m1mod2k: "- 1 mod 2 ^ n = (2 ^ n - 1 :: int)" by (rule zmod_minus1) simp | lemma m1mod2k: "- 1 mod 2 ^ n = (2 ^ n - 1 :: int)" by (rule zmod_minus1) simp | proof (prove)
goal (1 subgoal):
1. - 1 mod 2 ^ n = 2 ^ n - 1 | lemma m1mod2k: "- 1 mod 2 ^ n = (2 ^ n - 1 :: int)" | unnamed_thy_26 | More_Divides | 1 |
|
[] | lemma sb_inc_lem: "a + 2^k < 0 \<Longrightarrow> a + 2^k + 2^(Suc k) \<le> (a + 2^k) mod 2^(Suc k)"
for a :: int using int_mod_ge' [where n = "2 ^ (Suc k)" and b = "a + 2 ^ k"] by simp | lemma sb_inc_lem: "a + 2^k < 0 \<Longrightarrow> a + 2^k + 2^(Suc k) \<le> (a + 2^k) mod 2^(Suc k)"
for a :: int using int_mod_ge' [where n = "2 ^ (Suc k)" and b = "a + 2 ^ k"] by simp | proof (prove)
goal (1 subgoal):
1. a + 2 ^ k < 0 \<Longrightarrow> a + 2 ^ k + 2 ^ Suc k \<le> (a + 2 ^ k) mod 2 ^ Suc k proof (prove)
using this:
\<lbrakk>a + 2 ^ k < 0; 0 < 2 ^ Suc k\<rbrakk> \<Longrightarrow> a + 2 ^ k + 2 ^ Suc k \<le> (a + 2 ^ k) mod 2 ^ Suc k
goal (1 subgoal):
1. a + 2 ^ k < 0 \<Longrightarrow> a + 2 ^ k + 2 ^ Suc k \<le> (a + 2 ^ k) mod 2 ^ Suc k | lemma sb_inc_lem: "a + 2^k < 0 \<Longrightarrow> a + 2^k + 2^(Suc k) \<le> (a + 2^k) mod 2^(Suc k)"
for a :: int | unnamed_thy_27 | More_Divides | 2 |
|
[] | lemma sb_inc_lem': "a < - (2^k) \<Longrightarrow> a + 2^k + 2^(Suc k) \<le> (a + 2^k) mod 2^(Suc k)"
for a :: int by (rule sb_inc_lem) simp | lemma sb_inc_lem': "a < - (2^k) \<Longrightarrow> a + 2^k + 2^(Suc k) \<le> (a + 2^k) mod 2^(Suc k)"
for a :: int by (rule sb_inc_lem) simp | proof (prove)
goal (1 subgoal):
1. a < - (2 ^ k) \<Longrightarrow> a + 2 ^ k + 2 ^ Suc k \<le> (a + 2 ^ k) mod 2 ^ Suc k | lemma sb_inc_lem': "a < - (2^k) \<Longrightarrow> a + 2^k + 2^(Suc k) \<le> (a + 2^k) mod 2^(Suc k)"
for a :: int | unnamed_thy_28 | More_Divides | 1 |
|
[] | lemma sb_dec_lem: "0 \<le> - (2 ^ k) + a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a"
for a :: int using int_mod_le'[where n = "2 ^ (Suc k)" and b = "a + 2 ^ k"] by simp | lemma sb_dec_lem: "0 \<le> - (2 ^ k) + a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a"
for a :: int using int_mod_le'[where n = "2 ^ (Suc k)" and b = "a + 2 ^ k"] by simp | proof (prove)
goal (1 subgoal):
1. 0 \<le> - (2 ^ k) + a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a proof (prove)
using this:
0 \<le> a + 2 ^ k - 2 ^ Suc k \<Longrightarrow> (a + 2 ^ k) mod 2 ^ Suc k \<le> a + 2 ^ k - 2 ^ Suc k
goal (1 subgoal):
1. 0 \<le> - (2 ^ k) + a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a | lemma sb_dec_lem: "0 \<le> - (2 ^ k) + a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a"
for a :: int | unnamed_thy_29 | More_Divides | 2 |
|
[] | lemma sb_dec_lem': "2 ^ k \<le> a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a"
for a :: int by (rule sb_dec_lem) simp | lemma sb_dec_lem': "2 ^ k \<le> a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a"
for a :: int by (rule sb_dec_lem) simp | proof (prove)
goal (1 subgoal):
1. 2 ^ k \<le> a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a | lemma sb_dec_lem': "2 ^ k \<le> a \<Longrightarrow> (a + 2 ^ k) mod (2 * 2 ^ k) \<le> - (2 ^ k) + a"
for a :: int | unnamed_thy_30 | More_Divides | 1 |
|
[] | lemma mod_2_neq_1_eq_eq_0: "k mod 2 \<noteq> 1 \<longleftrightarrow> k mod 2 = 0"
for k :: int by (fact not_mod_2_eq_1_eq_0) | lemma mod_2_neq_1_eq_eq_0: "k mod 2 \<noteq> 1 \<longleftrightarrow> k mod 2 = 0"
for k :: int by (fact not_mod_2_eq_1_eq_0) | proof (prove)
goal (1 subgoal):
1. (k mod 2 \<noteq> 1) = (k mod 2 = 0) | lemma mod_2_neq_1_eq_eq_0: "k mod 2 \<noteq> 1 \<longleftrightarrow> k mod 2 = 0"
for k :: int | unnamed_thy_31 | More_Divides | 1 |
|
[] | lemma z1pmod2: "(2 * b + 1) mod 2 = (1::int)"
for b :: int by arith | lemma z1pmod2: "(2 * b + 1) mod 2 = (1::int)"
for b :: int by arith | proof (prove)
goal (1 subgoal):
1. (2 * b + 1) mod 2 = 1 | lemma z1pmod2: "(2 * b + 1) mod 2 = (1::int)"
for b :: int | unnamed_thy_32 | More_Divides | 1 |
|
[] | lemma p1mod22k': "(1 + 2 * b) mod (2 * 2 ^ n) = 1 + 2 * (b mod 2 ^ n)"
for b :: int by (rule pos_zmod_mult_2) simp | lemma p1mod22k': "(1 + 2 * b) mod (2 * 2 ^ n) = 1 + 2 * (b mod 2 ^ n)"
for b :: int by (rule pos_zmod_mult_2) simp | proof (prove)
goal (1 subgoal):
1. (1 + 2 * b) mod (2 * 2 ^ n) = 1 + 2 * (b mod 2 ^ n) | lemma p1mod22k': "(1 + 2 * b) mod (2 * 2 ^ n) = 1 + 2 * (b mod 2 ^ n)"
for b :: int | unnamed_thy_33 | More_Divides | 1 |
|
[] | lemma p1mod22k: "(2 * b + 1) mod (2 * 2 ^ n) = 2 * (b mod 2 ^ n) + 1"
for b :: int by (simp add: p1mod22k' add.commute) | lemma p1mod22k: "(2 * b + 1) mod (2 * 2 ^ n) = 2 * (b mod 2 ^ n) + 1"
for b :: int by (simp add: p1mod22k' add.commute) | proof (prove)
goal (1 subgoal):
1. (2 * b + 1) mod (2 * 2 ^ n) = 2 * (b mod 2 ^ n) + 1 | lemma p1mod22k: "(2 * b + 1) mod (2 * 2 ^ n) = 2 * (b mod 2 ^ n) + 1"
for b :: int | unnamed_thy_34 | More_Divides | 1 |
|
[] | lemma pos_mod_sign2:
\<open>0 \<le> a mod 2\<close> for a :: int by simp | lemma pos_mod_sign2:
\<open>0 \<le> a mod 2\<close> for a :: int by simp | proof (prove)
goal (1 subgoal):
1. 0 \<le> a mod 2 | lemma pos_mod_sign2:
\<open>0 \<le> a mod 2\<close> for a :: int | unnamed_thy_35 | More_Divides | 1 |
|
[] | lemma pos_mod_bound2:
\<open>a mod 2 < 2\<close> for a :: int by simp | lemma pos_mod_bound2:
\<open>a mod 2 < 2\<close> for a :: int by simp | proof (prove)
goal (1 subgoal):
1. a mod 2 < 2 | lemma pos_mod_bound2:
\<open>a mod 2 < 2\<close> for a :: int | unnamed_thy_36 | More_Divides | 1 |
|
[] | lemma nmod2: "n mod 2 = 0 \<or> n mod 2 = 1"
for n :: int by arith | lemma nmod2: "n mod 2 = 0 \<or> n mod 2 = 1"
for n :: int by arith | proof (prove)
goal (1 subgoal):
1. n mod 2 = 0 \<or> n mod 2 = 1 | lemma nmod2: "n mod 2 = 0 \<or> n mod 2 = 1"
for n :: int | unnamed_thy_37 | More_Divides | 1 |
|
[] | lemma eme1p:
"even n \<Longrightarrow> even d \<Longrightarrow> 0 \<le> d \<Longrightarrow> (1 + n) mod d = 1 + n mod d" for n d :: int using emep1 [of n d] by (simp add: ac_simps) | lemma eme1p:
"even n \<Longrightarrow> even d \<Longrightarrow> 0 \<le> d \<Longrightarrow> (1 + n) mod d = 1 + n mod d" for n d :: int using emep1 [of n d] by (simp add: ac_simps) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>even n; even d; 0 \<le> d\<rbrakk> \<Longrightarrow> (1 + n) mod d = 1 + n mod d proof (prove)
using this:
\<lbrakk>even n; even d; 0 \<le> d\<rbrakk> \<Longrightarrow> (n + 1) mod d = n mod d + 1
goal (1 subgoal):
1. \<lbrakk>even n; even d; 0 \<le> d\<rbrakk> \<Longrightarrow> (1 + n) mod d = 1 + n mod d | lemma eme1p:
"even n \<Longrightarrow> even d \<Longrightarrow> 0 \<le> d \<Longrightarrow> (1 + n) mod d = 1 + n mod d" for n d :: int | unnamed_thy_38 | More_Divides | 2 |
|
[] | lemma m1mod22k:
\<open>- 1 mod (2 * 2 ^ n) = 2 * 2 ^ n - (1::int)\<close> by (simp add: zmod_minus1) | lemma m1mod22k:
\<open>- 1 mod (2 * 2 ^ n) = 2 * 2 ^ n - (1::int)\<close> by (simp add: zmod_minus1) | proof (prove)
goal (1 subgoal):
1. - 1 mod (2 * 2 ^ n) = 2 * 2 ^ n - 1 | lemma m1mod22k:
\<open>- 1 mod (2 * 2 ^ n) = 2 * 2 ^ n - (1::int)\<close> | unnamed_thy_39 | More_Divides | 1 |
|
[] | lemma z1pdiv2: "(2 * b + 1) div 2 = b"
for b :: int by arith | lemma z1pdiv2: "(2 * b + 1) div 2 = b"
for b :: int by arith | proof (prove)
goal (1 subgoal):
1. (2 * b + 1) div 2 = b | lemma z1pdiv2: "(2 * b + 1) div 2 = b"
for b :: int | unnamed_thy_40 | More_Divides | 1 |
|
[] | lemma zdiv_le_dividend:
\<open>0 \<le> a \<Longrightarrow> 0 < b \<Longrightarrow> a div b \<le> a\<close> for a b :: int by (metis div_by_1 int_one_le_iff_zero_less zdiv_mono2 zero_less_one) | lemma zdiv_le_dividend:
\<open>0 \<le> a \<Longrightarrow> 0 < b \<Longrightarrow> a div b \<le> a\<close> for a b :: int by (metis div_by_1 int_one_le_iff_zero_less zdiv_mono2 zero_less_one) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>0 \<le> a; 0 < b\<rbrakk> \<Longrightarrow> a div b \<le> a | lemma zdiv_le_dividend:
\<open>0 \<le> a \<Longrightarrow> 0 < b \<Longrightarrow> a div b \<le> a\<close> for a b :: int | unnamed_thy_41 | More_Divides | 1 |
|
[] | lemma axxmod2: "(1 + x + x) mod 2 = 1 \<and> (0 + x + x) mod 2 = 0"
for x :: int by arith | lemma axxmod2: "(1 + x + x) mod 2 = 1 \<and> (0 + x + x) mod 2 = 0"
for x :: int by arith | proof (prove)
goal (1 subgoal):
1. (1 + x + x) mod 2 = 1 \<and> (0 + x + x) mod 2 = 0 | lemma axxmod2: "(1 + x + x) mod 2 = 1 \<and> (0 + x + x) mod 2 = 0"
for x :: int | unnamed_thy_42 | More_Divides | 1 |
|
[] | lemma axxdiv2: "(1 + x + x) div 2 = x \<and> (0 + x + x) div 2 = x"
for x :: int by arith | lemma axxdiv2: "(1 + x + x) div 2 = x \<and> (0 + x + x) div 2 = x"
for x :: int by arith | proof (prove)
goal (1 subgoal):
1. (1 + x + x) div 2 = x \<and> (0 + x + x) div 2 = x | lemma axxdiv2: "(1 + x + x) div 2 = x \<and> (0 + x + x) div 2 = x"
for x :: int | unnamed_thy_43 | More_Divides | 1 |
|
[] | lemma mod_plus_right: "(a + x) mod m = (b + x) mod m \<longleftrightarrow> a mod m = b mod m"
for a b m x :: nat by (induct x) (simp_all add: mod_Suc, arith) | lemma mod_plus_right: "(a + x) mod m = (b + x) mod m \<longleftrightarrow> a mod m = b mod m"
for a b m x :: nat by (induct x) (simp_all add: mod_Suc, arith) | proof (prove)
goal (1 subgoal):
1. ((a + x) mod m = (b + x) mod m) = (a mod m = b mod m) | lemma mod_plus_right: "(a + x) mod m = (b + x) mod m \<longleftrightarrow> a mod m = b mod m"
for a b m x :: nat | unnamed_thy_44 | More_Divides | 1 |
|
[] | lemma nat_minus_mod: "(n - n mod m) mod m = 0"
for m n :: nat by (induct n) (simp_all add: mod_Suc) | lemma nat_minus_mod: "(n - n mod m) mod m = 0"
for m n :: nat by (induct n) (simp_all add: mod_Suc) | proof (prove)
goal (1 subgoal):
1. (n - n mod m) mod m = 0 | lemma nat_minus_mod: "(n - n mod m) mod m = 0"
for m n :: nat | unnamed_thy_45 | More_Divides | 1 |
|
[] | lemma nat_mod_eq: "b < n \<Longrightarrow> a mod n = b mod n \<Longrightarrow> a mod n = b"
for a b n :: nat by (induct a) auto | lemma nat_mod_eq: "b < n \<Longrightarrow> a mod n = b mod n \<Longrightarrow> a mod n = b"
for a b n :: nat by (induct a) auto | proof (prove)
goal (1 subgoal):
1. \<lbrakk>b < n; a mod n = b mod n\<rbrakk> \<Longrightarrow> a mod n = b | lemma nat_mod_eq: "b < n \<Longrightarrow> a mod n = b mod n \<Longrightarrow> a mod n = b"
for a b n :: nat | unnamed_thy_46 | More_Divides | 1 |
|
[] | lemma mod_nat_add: "x < z \<Longrightarrow> y < z \<Longrightarrow> (x + y) mod z = (if x + y < z then x + y else x + y - z)"
for x y z :: nat apply (rule nat_mod_eq) apply auto apply (rule trans) apply (rule le_mod_geq) apply simp apply (rule nat_mod_eq') apply arith done | lemma mod_nat_add: "x < z \<Longrightarrow> y < z \<Longrightarrow> (x + y) mod z = (if x + y < z then x + y else x + y - z)"
for x y z :: nat apply (rule nat_mod_eq) apply auto apply (rule trans) apply (rule le_mod_geq) apply simp apply (rule nat_mod_eq') apply arith done | proof (prove)
goal (1 subgoal):
1. \<lbrakk>x < z; y < z\<rbrakk> \<Longrightarrow> (x + y) mod z = (if x + y < z then x + y else x + y - z) proof (prove)
goal (2 subgoals):
1. \<lbrakk>x < z; y < z\<rbrakk> \<Longrightarrow> (if x + y < z then x + y else x + y - z) < z
2. \<lbrakk>x < z; y < z\<rbrakk> \<Longrightarrow> (x + y) mod z = (if x + y < z then x + y else x + y - z) mod z proof (prove)
goal (1 subgoal):
1. \<lbrakk>x < z; y < z; \<not> x + y < z\<rbrakk> \<Longrightarrow> (x + y) mod z = x + y - z proof (prove)
goal (2 subgoals):
1. \<lbrakk>x < z; y < z; \<not> x + y < z\<rbrakk> \<Longrightarrow> (x + y) mod z = ?s56
2. \<lbrakk>x < z; y < z; \<not> x + y < z\<rbrakk> \<Longrightarrow> ?s56 = x + y - z proof (prove)
goal (2 subgoals):
1. \<lbrakk>x < z; y < z; \<not> x + y < z\<rbrakk> \<Longrightarrow> z \<le> x + y
2. \<lbrakk>x < z; y < z; \<not> x + y < z\<rbrakk> \<Longrightarrow> (x + y - z) mod z = x + y - z proof (prove)
goal (1 subgoal):
1. \<lbrakk>x < z; y < z; \<not> x + y < z\<rbrakk> \<Longrightarrow> (x + y - z) mod z = x + y - z proof (prove)
goal (1 subgoal):
1. \<lbrakk>x < z; y < z; \<not> x + y < z\<rbrakk> \<Longrightarrow> x + y - z < z proof (prove)
goal:
No subgoals! | lemma mod_nat_add: "x < z \<Longrightarrow> y < z \<Longrightarrow> (x + y) mod z = (if x + y < z then x + y else x + y - z)"
for x y z :: nat | unnamed_thy_48 | More_Divides | 8 |
|
[] | lemma mod_nat_sub: "x < z \<Longrightarrow> (x - y) mod z = x - y"
for x y :: nat by (rule nat_mod_eq') arith | lemma mod_nat_sub: "x < z \<Longrightarrow> (x - y) mod z = x - y"
for x y :: nat by (rule nat_mod_eq') arith | proof (prove)
goal (1 subgoal):
1. x < z \<Longrightarrow> (x - y) mod z = x - y | lemma mod_nat_sub: "x < z \<Longrightarrow> (x - y) mod z = x - y"
for x y :: nat | unnamed_thy_49 | More_Divides | 1 |
|
[] | lemma int_mod_eq: "0 \<le> b \<Longrightarrow> b < n \<Longrightarrow> a mod n = b mod n \<Longrightarrow> a mod n = b"
for a b n :: int by (metis mod_pos_pos_trivial) | lemma int_mod_eq: "0 \<le> b \<Longrightarrow> b < n \<Longrightarrow> a mod n = b mod n \<Longrightarrow> a mod n = b"
for a b n :: int by (metis mod_pos_pos_trivial) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>0 \<le> b; b < n; a mod n = b mod n\<rbrakk> \<Longrightarrow> a mod n = b | lemma int_mod_eq: "0 \<le> b \<Longrightarrow> b < n \<Longrightarrow> a mod n = b mod n \<Longrightarrow> a mod n = b"
for a b n :: int | unnamed_thy_50 | More_Divides | 1 |
|
[] | lemma zmde:
\<open>b * (a div b) = a - a mod b\<close> for a b :: \<open>'a::{group_add,semiring_modulo}\<close> using mult_div_mod_eq [of b a] by (simp add: eq_diff_eq) | lemma zmde:
\<open>b * (a div b) = a - a mod b\<close> for a b :: \<open>'a::{group_add,semiring_modulo}\<close> using mult_div_mod_eq [of b a] by (simp add: eq_diff_eq) | proof (prove)
goal (1 subgoal):
1. b * (a div b) = a - a mod b proof (prove)
using this:
b * (a div b) + a mod b = a
goal (1 subgoal):
1. b * (a div b) = a - a mod b | lemma zmde:
\<open>b * (a div b) = a - a mod b\<close> for a b :: \<open>'a::{group_add,semiring_modulo}\<close> | unnamed_thy_51 | More_Divides | 2 |
|
[] | lemma zdiv_mult_self: "m \<noteq> 0 \<Longrightarrow> (a + m * n) div m = a div m + n"
for a m n :: int by simp | lemma zdiv_mult_self: "m \<noteq> 0 \<Longrightarrow> (a + m * n) div m = a div m + n"
for a m n :: int by simp | proof (prove)
goal (1 subgoal):
1. m \<noteq> 0 \<Longrightarrow> (a + m * n) div m = a div m + n | lemma zdiv_mult_self: "m \<noteq> 0 \<Longrightarrow> (a + m * n) div m = a div m + n"
for a m n :: int | unnamed_thy_52 | More_Divides | 1 |
|
[] | lemma mod_power_lem: "a > 1 \<Longrightarrow> a ^ n mod a ^ m = (if m \<le> n then 0 else a ^ n)"
for a :: int by (simp add: mod_eq_0_iff_dvd le_imp_power_dvd) | lemma mod_power_lem: "a > 1 \<Longrightarrow> a ^ n mod a ^ m = (if m \<le> n then 0 else a ^ n)"
for a :: int by (simp add: mod_eq_0_iff_dvd le_imp_power_dvd) | proof (prove)
goal (1 subgoal):
1. 1 < a \<Longrightarrow> a ^ n mod a ^ m = (if m \<le> n then 0 else a ^ n) | lemma mod_power_lem: "a > 1 \<Longrightarrow> a ^ n mod a ^ m = (if m \<le> n then 0 else a ^ n)"
for a :: int | unnamed_thy_53 | More_Divides | 1 |
|
[] | lemma nonneg_mod_div: "0 \<le> a \<Longrightarrow> 0 \<le> b \<Longrightarrow> 0 \<le> (a mod b) \<and> 0 \<le> a div b"
for a b :: int by (cases "b = 0") (auto intro: pos_imp_zdiv_nonneg_iff [THEN iffD2]) | lemma nonneg_mod_div: "0 \<le> a \<Longrightarrow> 0 \<le> b \<Longrightarrow> 0 \<le> (a mod b) \<and> 0 \<le> a div b"
for a b :: int by (cases "b = 0") (auto intro: pos_imp_zdiv_nonneg_iff [THEN iffD2]) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>0 \<le> a; 0 \<le> b\<rbrakk> \<Longrightarrow> 0 \<le> a mod b \<and> 0 \<le> a div b | lemma nonneg_mod_div: "0 \<le> a \<Longrightarrow> 0 \<le> b \<Longrightarrow> 0 \<le> (a mod b) \<and> 0 \<le> a div b"
for a b :: int | unnamed_thy_54 | More_Divides | 1 |
|
[] | lemma mod_exp_less_eq_exp:
\<open>a mod 2 ^ n < 2 ^ n\<close> for a :: int by (rule pos_mod_bound) simp | lemma mod_exp_less_eq_exp:
\<open>a mod 2 ^ n < 2 ^ n\<close> for a :: int by (rule pos_mod_bound) simp | proof (prove)
goal (1 subgoal):
1. a mod 2 ^ n < 2 ^ n | lemma mod_exp_less_eq_exp:
\<open>a mod 2 ^ n < 2 ^ n\<close> for a :: int | unnamed_thy_55 | More_Divides | 1 |
|
[] | lemma div_mult_le:
\<open>a div b * b \<le> a\<close> for a b :: nat by (fact div_times_less_eq_dividend) | lemma div_mult_le:
\<open>a div b * b \<le> a\<close> for a b :: nat by (fact div_times_less_eq_dividend) | proof (prove)
goal (1 subgoal):
1. a div b * b \<le> a | lemma div_mult_le:
\<open>a div b * b \<le> a\<close> for a b :: nat | unnamed_thy_56 | More_Divides | 1 |
|
[] | lemma power_sub:
fixes a :: nat
assumes lt: "n \<le> m"
and av: "0 < a"
shows "a ^ (m - n) = a ^ m div a ^ n" proof (subst nat_mult_eq_cancel1 [symmetric]) show "(0::nat) < a ^ n" using av by simp next from lt obtain q where mv: "n + q = m" by (auto simp: le_iff_add) have "a ^ n * (a ^ m div a ^ n) = a ^ m" proof (subst mult.commute) have "a ^ m = (a ^ m div a ^ n) * a ^ n + a ^ m mod a ^ n" by (rule div_mult_mod_eq [symmetric]) moreover have "a ^ m mod a ^ n = 0" by (subst mod_eq_0_iff_dvd, subst dvd_def, rule exI [where x = "a ^ q"],
(subst power_add [symmetric] mv)+, rule refl) ultimately show "(a ^ m div a ^ n) * a ^ n = a ^ m" by simp qed then show "a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n)" using lt by (simp add: power_add [symmetric]) qed | lemma power_sub:
fixes a :: nat
assumes lt: "n \<le> m"
and av: "0 < a"
shows "a ^ (m - n) = a ^ m div a ^ n" proof (subst nat_mult_eq_cancel1 [symmetric]) show "(0::nat) < a ^ n" using av by simp next from lt obtain q where mv: "n + q = m" by (auto simp: le_iff_add) have "a ^ n * (a ^ m div a ^ n) = a ^ m" proof (subst mult.commute) have "a ^ m = (a ^ m div a ^ n) * a ^ n + a ^ m mod a ^ n" by (rule div_mult_mod_eq [symmetric]) moreover have "a ^ m mod a ^ n = 0" by (subst mod_eq_0_iff_dvd, subst dvd_def, rule exI [where x = "a ^ q"],
(subst power_add [symmetric] mv)+, rule refl) ultimately show "(a ^ m div a ^ n) * a ^ n = a ^ m" by simp qed then show "a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n)" using lt by (simp add: power_add [symmetric]) qed | proof (prove)
goal (1 subgoal):
1. a ^ (m - n) = a ^ m div a ^ n proof (state)
goal (2 subgoals):
1. 0 < ?k
2. ?k * a ^ (m - n) = ?k * (a ^ m div a ^ n) proof (prove)
goal (1 subgoal):
1. 0 < a ^ n proof (prove)
using this:
0 < a
goal (1 subgoal):
1. 0 < a ^ n proof (state)
this:
0 < a ^ n
goal (1 subgoal):
1. a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n) proof (state)
goal (1 subgoal):
1. a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n) proof (chain)
picking this:
n \<le> m proof (prove)
using this:
n \<le> m
goal (1 subgoal):
1. (\<And>q. n + q = m \<Longrightarrow> thesis) \<Longrightarrow> thesis proof (state)
this:
n + q = m
goal (1 subgoal):
1. a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n) proof (prove)
goal (1 subgoal):
1. a ^ n * (a ^ m div a ^ n) = a ^ m proof (state)
goal (1 subgoal):
1. a ^ m div a ^ n * a ^ n = a ^ m proof (prove)
goal (1 subgoal):
1. a ^ m = a ^ m div a ^ n * a ^ n + a ^ m mod a ^ n proof (state)
this:
a ^ m = a ^ m div a ^ n * a ^ n + a ^ m mod a ^ n
goal (1 subgoal):
1. a ^ m div a ^ n * a ^ n = a ^ m proof (state)
this:
a ^ m = a ^ m div a ^ n * a ^ n + a ^ m mod a ^ n
goal (1 subgoal):
1. a ^ m div a ^ n * a ^ n = a ^ m proof (prove)
goal (1 subgoal):
1. a ^ m mod a ^ n = 0 proof (state)
this:
a ^ m mod a ^ n = 0
goal (1 subgoal):
1. a ^ m div a ^ n * a ^ n = a ^ m proof (chain)
picking this:
a ^ m = a ^ m div a ^ n * a ^ n + a ^ m mod a ^ n
a ^ m mod a ^ n = 0 proof (prove)
using this:
a ^ m = a ^ m div a ^ n * a ^ n + a ^ m mod a ^ n
a ^ m mod a ^ n = 0
goal (1 subgoal):
1. a ^ m div a ^ n * a ^ n = a ^ m proof (state)
this:
a ^ m div a ^ n * a ^ n = a ^ m
goal:
No subgoals! proof (state)
this:
a ^ n * (a ^ m div a ^ n) = a ^ m
goal (1 subgoal):
1. a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n) proof (chain)
picking this:
a ^ n * (a ^ m div a ^ n) = a ^ m proof (prove)
using this:
a ^ n * (a ^ m div a ^ n) = a ^ m
goal (1 subgoal):
1. a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n) proof (prove)
using this:
a ^ n * (a ^ m div a ^ n) = a ^ m
n \<le> m
goal (1 subgoal):
1. a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n) proof (state)
this:
a ^ n * a ^ (m - n) = a ^ n * (a ^ m div a ^ n)
goal:
No subgoals! | lemma power_sub:
fixes a :: nat
assumes lt: "n \<le> m"
and av: "0 < a"
shows "a ^ (m - n) = a ^ m div a ^ n" | unnamed_thy_57 | More_Divides | 24 |
|
[] | lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c" apply (cut_tac m = q and n = c in mod_less_divisor) apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto) apply (erule_tac P = "%x. lhs < rhs x" for lhs rhs in ssubst) apply (simp add: add_mult_distrib2) done | lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c" apply (cut_tac m = q and n = c in mod_less_divisor) apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto) apply (erule_tac P = "%x. lhs < rhs x" for lhs rhs in ssubst) apply (simp add: add_mult_distrib2) done | proof (prove)
goal (1 subgoal):
1. \<lbrakk>0 < c; r < b\<rbrakk> \<Longrightarrow> b * (q mod c) + r < b * c proof (prove)
goal (2 subgoals):
1. \<lbrakk>0 < c; r < b\<rbrakk> \<Longrightarrow> 0 < c
2. \<lbrakk>0 < c; r < b; q mod c < c\<rbrakk> \<Longrightarrow> b * (q mod c) + r < b * c proof (prove)
goal (1 subgoal):
1. \<And>k. \<lbrakk>r < b; c = Suc (q mod c + k)\<rbrakk> \<Longrightarrow> b * (q mod c) + r < b * c proof (prove)
goal (1 subgoal):
1. \<And>k. r < b \<Longrightarrow> b * (q mod c) + r < b * Suc (q mod c + k) proof (prove)
goal:
No subgoals! | lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c" | unnamed_thy_58 | More_Divides | 5 |
|
[] | lemma less_two_pow_divD:
"\<lbrakk> (x :: nat) < 2 ^ n div 2 ^ m \<rbrakk>
\<Longrightarrow> n \<ge> m \<and> (x < 2 ^ (n - m))" apply (rule context_conjI) apply (rule ccontr) apply (simp add: power_strict_increasing) apply (simp add: power_sub) done | lemma less_two_pow_divD:
"\<lbrakk> (x :: nat) < 2 ^ n div 2 ^ m \<rbrakk>
\<Longrightarrow> n \<ge> m \<and> (x < 2 ^ (n - m))" apply (rule context_conjI) apply (rule ccontr) apply (simp add: power_strict_increasing) apply (simp add: power_sub) done | proof (prove)
goal (1 subgoal):
1. x < 2 ^ n div 2 ^ m \<Longrightarrow> m \<le> n \<and> x < 2 ^ (n - m) proof (prove)
goal (2 subgoals):
1. x < 2 ^ n div 2 ^ m \<Longrightarrow> m \<le> n
2. \<lbrakk>x < 2 ^ n div 2 ^ m; m \<le> n\<rbrakk> \<Longrightarrow> x < 2 ^ (n - m) proof (prove)
goal (2 subgoals):
1. \<lbrakk>x < 2 ^ n div 2 ^ m; \<not> m \<le> n\<rbrakk> \<Longrightarrow> False
2. \<lbrakk>x < 2 ^ n div 2 ^ m; m \<le> n\<rbrakk> \<Longrightarrow> x < 2 ^ (n - m) proof (prove)
goal (1 subgoal):
1. \<lbrakk>x < 2 ^ n div 2 ^ m; m \<le> n\<rbrakk> \<Longrightarrow> x < 2 ^ (n - m) proof (prove)
goal:
No subgoals! | lemma less_two_pow_divD:
"\<lbrakk> (x :: nat) < 2 ^ n div 2 ^ m \<rbrakk>
\<Longrightarrow> n \<ge> m \<and> (x < 2 ^ (n - m))" | unnamed_thy_59 | More_Divides | 5 |
|
[] | lemma less_two_pow_divI:
"\<lbrakk> (x :: nat) < 2 ^ (n - m); m \<le> n \<rbrakk> \<Longrightarrow> x < 2 ^ n div 2 ^ m" by (simp add: power_sub) | lemma less_two_pow_divI:
"\<lbrakk> (x :: nat) < 2 ^ (n - m); m \<le> n \<rbrakk> \<Longrightarrow> x < 2 ^ n div 2 ^ m" by (simp add: power_sub) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>x < 2 ^ (n - m); m \<le> n\<rbrakk> \<Longrightarrow> x < 2 ^ n div 2 ^ m | lemma less_two_pow_divI:
"\<lbrakk> (x :: nat) < 2 ^ (n - m); m \<le> n \<rbrakk> \<Longrightarrow> x < 2 ^ n div 2 ^ m" | unnamed_thy_60 | More_Divides | 1 |
|
[] | lemma power_mod_div:
fixes x :: "nat"
shows "x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)" (is "?LHS = ?RHS") proof (cases "n \<le> m") case True then have "?LHS = 0" apply - apply (rule div_less) apply (rule order_less_le_trans [OF mod_less_divisor]; simp) done also have "\<dots> = ?RHS" using True by simp finally show ?thesis . next case False then have lt: "m < n" by simp then obtain q where nv: "n = m + q" and "0 < q" by (auto dest: less_imp_Suc_add) then have "x mod 2 ^ n = 2 ^ m * (x div 2 ^ m mod 2 ^ q) + x mod 2 ^ m" by (simp add: power_add mod_mult2_eq) then have "?LHS = x div 2 ^ m mod 2 ^ q" by (simp add: div_add1_eq) also have "\<dots> = ?RHS" using nv by simp finally show ?thesis . qed | lemma power_mod_div:
fixes x :: "nat"
shows "x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)" (is "?LHS = ?RHS") proof (cases "n \<le> m") case True then have "?LHS = 0" apply - apply (rule div_less) apply (rule order_less_le_trans [OF mod_less_divisor]; simp) done also have "\<dots> = ?RHS" using True by simp finally show ?thesis . next case False then have lt: "m < n" by simp then obtain q where nv: "n = m + q" and "0 < q" by (auto dest: less_imp_Suc_add) then have "x mod 2 ^ n = 2 ^ m * (x div 2 ^ m mod 2 ^ q) + x mod 2 ^ m" by (simp add: power_add mod_mult2_eq) then have "?LHS = x div 2 ^ m mod 2 ^ q" by (simp add: div_add1_eq) also have "\<dots> = ?RHS" using nv by simp finally show ?thesis . qed | proof (prove)
goal (1 subgoal):
1. x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (state)
goal (2 subgoals):
1. n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
2. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (state)
this:
n \<le> m
goal (2 subgoals):
1. n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
2. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (chain)
picking this:
n \<le> m proof (prove)
using this:
n \<le> m
goal (1 subgoal):
1. x mod 2 ^ n div 2 ^ m = 0 proof (prove)
goal (1 subgoal):
1. n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = 0 proof (prove)
goal (1 subgoal):
1. n \<le> m \<Longrightarrow> x mod 2 ^ n < 2 ^ m proof (prove)
goal:
No subgoals! proof (state)
this:
x mod 2 ^ n div 2 ^ m = 0
goal (2 subgoals):
1. n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
2. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (state)
this:
x mod 2 ^ n div 2 ^ m = 0
goal (2 subgoals):
1. n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
2. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (prove)
goal (1 subgoal):
1. 0 = x div 2 ^ m mod 2 ^ (n - m) proof (prove)
using this:
n \<le> m
goal (1 subgoal):
1. 0 = x div 2 ^ m mod 2 ^ (n - m) proof (state)
this:
0 = x div 2 ^ m mod 2 ^ (n - m)
goal (2 subgoals):
1. n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
2. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (chain)
picking this:
x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (prove)
using this:
x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
goal (1 subgoal):
1. x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (state)
this:
x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (state)
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (state)
this:
\<not> n \<le> m
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (chain)
picking this:
\<not> n \<le> m proof (prove)
using this:
\<not> n \<le> m
goal (1 subgoal):
1. m < n proof (state)
this:
m < n
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (chain)
picking this:
m < n proof (prove)
using this:
m < n
goal (1 subgoal):
1. (\<And>q. \<lbrakk>n = m + q; 0 < q\<rbrakk> \<Longrightarrow> thesis) \<Longrightarrow> thesis proof (state)
this:
n = m + q
0 < q
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (chain)
picking this:
n = m + q
0 < q proof (prove)
using this:
n = m + q
0 < q
goal (1 subgoal):
1. x mod 2 ^ n = 2 ^ m * (x div 2 ^ m mod 2 ^ q) + x mod 2 ^ m proof (state)
this:
x mod 2 ^ n = 2 ^ m * (x div 2 ^ m mod 2 ^ q) + x mod 2 ^ m
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (chain)
picking this:
x mod 2 ^ n = 2 ^ m * (x div 2 ^ m mod 2 ^ q) + x mod 2 ^ m proof (prove)
using this:
x mod 2 ^ n = 2 ^ m * (x div 2 ^ m mod 2 ^ q) + x mod 2 ^ m
goal (1 subgoal):
1. x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ q proof (state)
this:
x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ q
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (state)
this:
x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ q
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (prove)
goal (1 subgoal):
1. x div 2 ^ m mod 2 ^ q = x div 2 ^ m mod 2 ^ (n - m) proof (prove)
using this:
n = m + q
goal (1 subgoal):
1. x div 2 ^ m mod 2 ^ q = x div 2 ^ m mod 2 ^ (n - m) proof (state)
this:
x div 2 ^ m mod 2 ^ q = x div 2 ^ m mod 2 ^ (n - m)
goal (1 subgoal):
1. \<not> n \<le> m \<Longrightarrow> x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (chain)
picking this:
x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (prove)
using this:
x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
goal (1 subgoal):
1. x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m) proof (state)
this:
x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)
goal:
No subgoals! | lemma power_mod_div:
fixes x :: "nat"
shows "x mod 2 ^ n div 2 ^ m = x div 2 ^ m mod 2 ^ (n - m)" (is "?LHS = ?RHS") | unnamed_thy_61 | More_Divides | 37 |
|
[] | lemma mod_mod_power:
fixes k :: nat
shows "k mod 2 ^ m mod 2 ^ n = k mod 2 ^ (min m n)" proof (cases "m \<le> n") case True then have "k mod 2 ^ m mod 2 ^ n = k mod 2 ^ m" apply - apply (subst mod_less [where n = "2 ^ n"]) apply (rule order_less_le_trans [OF mod_less_divisor]) apply simp+ done also have "\<dots> = k mod 2 ^ (min m n)" using True by simp finally show ?thesis . next case False then have "n < m" by simp then obtain d where md: "m = n + d" by (auto dest: less_imp_add_positive) then have "k mod 2 ^ m = 2 ^ n * (k div 2 ^ n mod 2 ^ d) + k mod 2 ^ n" by (simp add: mod_mult2_eq power_add) then have "k mod 2 ^ m mod 2 ^ n = k mod 2 ^ n" by (simp add: mod_add_left_eq) then show ?thesis using False by simp qed | lemma mod_mod_power:
fixes k :: nat
shows "k mod 2 ^ m mod 2 ^ n = k mod 2 ^ (min m n)" proof (cases "m \<le> n") case True then have "k mod 2 ^ m mod 2 ^ n = k mod 2 ^ m" apply - apply (subst mod_less [where n = "2 ^ n"]) apply (rule order_less_le_trans [OF mod_less_divisor]) apply simp+ done also have "\<dots> = k mod 2 ^ (min m n)" using True by simp finally show ?thesis . next case False then have "n < m" by simp then obtain d where md: "m = n + d" by (auto dest: less_imp_add_positive) then have "k mod 2 ^ m = 2 ^ n * (k div 2 ^ n mod 2 ^ d) + k mod 2 ^ n" by (simp add: mod_mult2_eq power_add) then have "k mod 2 ^ m mod 2 ^ n = k mod 2 ^ n" by (simp add: mod_add_left_eq) then show ?thesis using False by simp qed | proof (prove)
goal (1 subgoal):
1. k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (state)
goal (2 subgoals):
1. m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n
2. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (state)
this:
m \<le> n
goal (2 subgoals):
1. m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n
2. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (chain)
picking this:
m \<le> n proof (prove)
using this:
m \<le> n
goal (1 subgoal):
1. k mod 2 ^ m mod 2 ^ n = k mod 2 ^ m proof (prove)
goal (1 subgoal):
1. m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ m proof (prove)
goal (2 subgoals):
1. m \<le> n \<Longrightarrow> k mod 2 ^ m < 2 ^ n
2. m \<le> n \<Longrightarrow> k mod 2 ^ m = k mod 2 ^ m proof (prove)
goal (3 subgoals):
1. m \<le> n \<Longrightarrow> 0 < 2 ^ m
2. m \<le> n \<Longrightarrow> 2 ^ m \<le> 2 ^ n
3. m \<le> n \<Longrightarrow> k mod 2 ^ m = k mod 2 ^ m proof (prove)
goal:
No subgoals! proof (state)
this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ m
goal (2 subgoals):
1. m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n
2. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (state)
this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ m
goal (2 subgoals):
1. m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n
2. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (prove)
goal (1 subgoal):
1. k mod 2 ^ m = k mod 2 ^ min m n proof (prove)
using this:
m \<le> n
goal (1 subgoal):
1. k mod 2 ^ m = k mod 2 ^ min m n proof (state)
this:
k mod 2 ^ m = k mod 2 ^ min m n
goal (2 subgoals):
1. m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n
2. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (chain)
picking this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (prove)
using this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n
goal (1 subgoal):
1. k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (state)
this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n
goal (1 subgoal):
1. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (state)
goal (1 subgoal):
1. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (state)
this:
\<not> m \<le> n
goal (1 subgoal):
1. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (chain)
picking this:
\<not> m \<le> n proof (prove)
using this:
\<not> m \<le> n
goal (1 subgoal):
1. n < m proof (state)
this:
n < m
goal (1 subgoal):
1. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (chain)
picking this:
n < m proof (prove)
using this:
n < m
goal (1 subgoal):
1. (\<And>d. m = n + d \<Longrightarrow> thesis) \<Longrightarrow> thesis proof (state)
this:
m = n + d
goal (1 subgoal):
1. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (chain)
picking this:
m = n + d proof (prove)
using this:
m = n + d
goal (1 subgoal):
1. k mod 2 ^ m = 2 ^ n * (k div 2 ^ n mod 2 ^ d) + k mod 2 ^ n proof (state)
this:
k mod 2 ^ m = 2 ^ n * (k div 2 ^ n mod 2 ^ d) + k mod 2 ^ n
goal (1 subgoal):
1. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (chain)
picking this:
k mod 2 ^ m = 2 ^ n * (k div 2 ^ n mod 2 ^ d) + k mod 2 ^ n proof (prove)
using this:
k mod 2 ^ m = 2 ^ n * (k div 2 ^ n mod 2 ^ d) + k mod 2 ^ n
goal (1 subgoal):
1. k mod 2 ^ m mod 2 ^ n = k mod 2 ^ n proof (state)
this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ n
goal (1 subgoal):
1. \<not> m \<le> n \<Longrightarrow> k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (chain)
picking this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ n proof (prove)
using this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ n
goal (1 subgoal):
1. k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (prove)
using this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ n
\<not> m \<le> n
goal (1 subgoal):
1. k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n proof (state)
this:
k mod 2 ^ m mod 2 ^ n = k mod 2 ^ min m n
goal:
No subgoals! | lemma mod_mod_power:
fixes k :: nat
shows "k mod 2 ^ m mod 2 ^ n = k mod 2 ^ (min m n)" | unnamed_thy_62 | More_Divides | 35 |
|
[] | lemma mod_div_equality_div_eq:
"a div b * b = (a - (a mod b) :: int)" by (simp add: field_simps) | lemma mod_div_equality_div_eq:
"a div b * b = (a - (a mod b) :: int)" by (simp add: field_simps) | proof (prove)
goal (1 subgoal):
1. a div b * b = a - a mod b | lemma mod_div_equality_div_eq:
"a div b * b = (a - (a mod b) :: int)" | unnamed_thy_63 | More_Divides | 1 |
|
[] | lemma zmod_helper:
"n mod m = k \<Longrightarrow> ((n :: int) + a) mod m = (k + a) mod m" by (metis add.commute mod_add_right_eq) | lemma zmod_helper:
"n mod m = k \<Longrightarrow> ((n :: int) + a) mod m = (k + a) mod m" by (metis add.commute mod_add_right_eq) | proof (prove)
goal (1 subgoal):
1. n mod m = k \<Longrightarrow> (n + a) mod m = (k + a) mod m | lemma zmod_helper:
"n mod m = k \<Longrightarrow> ((n :: int) + a) mod m = (k + a) mod m" | unnamed_thy_64 | More_Divides | 1 |
|
[] | lemma int_div_sub_1:
"\<lbrakk> m \<ge> 1 \<rbrakk> \<Longrightarrow> (n - (1 :: int)) div m = (if m dvd n then (n div m) - 1 else n div m)" apply (subgoal_tac "m = 0 \<or> (n - (1 :: int)) div m = (if m dvd n then (n div m) - 1 else n div m)") apply fastforce apply (subst mult_cancel_right[symmetric]) apply (simp only: left_diff_distrib split: if_split) apply (simp only: mod_div_equality_div_eq) apply (clarsimp simp: field_simps) apply (clarsimp simp: dvd_eq_mod_eq_0) apply (cases "m = 1") apply simp apply (subst mod_diff_eq[symmetric], simp add: zmod_minus1) apply clarsimp apply (subst diff_add_cancel[where b=1, symmetric]) apply (subst mod_add_eq[symmetric]) apply (simp add: field_simps) apply (rule mod_pos_pos_trivial) apply (subst add_0_right[where a=0, symmetric]) apply (rule add_mono) apply simp apply simp apply (cases "(n - 1) mod m = m - 1") apply (drule zmod_helper[where a=1]) apply simp apply (subgoal_tac "1 + (n - 1) mod m \<le> m") apply simp apply (subst field_simps, rule zless_imp_add1_zle) apply simp done | lemma int_div_sub_1:
"\<lbrakk> m \<ge> 1 \<rbrakk> \<Longrightarrow> (n - (1 :: int)) div m = (if m dvd n then (n div m) - 1 else n div m)" apply (subgoal_tac "m = 0 \<or> (n - (1 :: int)) div m = (if m dvd n then (n div m) - 1 else n div m)") apply fastforce apply (subst mult_cancel_right[symmetric]) apply (simp only: left_diff_distrib split: if_split) apply (simp only: mod_div_equality_div_eq) apply (clarsimp simp: field_simps) apply (clarsimp simp: dvd_eq_mod_eq_0) apply (cases "m = 1") apply simp apply (subst mod_diff_eq[symmetric], simp add: zmod_minus1) apply clarsimp apply (subst diff_add_cancel[where b=1, symmetric]) apply (subst mod_add_eq[symmetric]) apply (simp add: field_simps) apply (rule mod_pos_pos_trivial) apply (subst add_0_right[where a=0, symmetric]) apply (rule add_mono) apply simp apply simp apply (cases "(n - 1) mod m = m - 1") apply (drule zmod_helper[where a=1]) apply simp apply (subgoal_tac "1 + (n - 1) mod m \<le> m") apply simp apply (subst field_simps, rule zless_imp_add1_zle) apply simp done | proof (prove)
goal (1 subgoal):
1. 1 \<le> m \<Longrightarrow> (n - 1) div m = (if m dvd n then n div m - 1 else n div m) proof (prove)
goal (2 subgoals):
1. \<lbrakk>1 \<le> m; m = 0 \<or> (n - 1) div m = (if m dvd n then n div m - 1 else n div m)\<rbrakk> \<Longrightarrow> (n - 1) div m = (if m dvd n then n div m - 1 else n div m)
2. 1 \<le> m \<Longrightarrow> m = 0 \<or> (n - 1) div m = (if m dvd n then n div m - 1 else n div m) proof (prove)
goal (1 subgoal):
1. 1 \<le> m \<Longrightarrow> m = 0 \<or> (n - 1) div m = (if m dvd n then n div m - 1 else n div m) proof (prove)
goal (1 subgoal):
1. 1 \<le> m \<Longrightarrow> (n - 1) div m * m = (if m dvd n then n div m - 1 else n div m) * m proof (prove)
goal (1 subgoal):
1. 1 \<le> m \<Longrightarrow> (m dvd n \<longrightarrow> (n - 1) div m * m = n div m * m - 1 * m) \<and> (\<not> m dvd n \<longrightarrow> (n - 1) div m * m = n div m * m) proof (prove)
goal (1 subgoal):
1. 1 \<le> m \<Longrightarrow> (m dvd n \<longrightarrow> n - 1 - (n - 1) mod m = n - n mod m - 1 * m) \<and> (\<not> m dvd n \<longrightarrow> n - 1 - (n - 1) mod m = n - n mod m) proof (prove)
goal (1 subgoal):
1. 1 \<le> m \<Longrightarrow> (m dvd n \<longrightarrow> m = 1 + (n - 1) mod m) \<and> (\<not> m dvd n \<longrightarrow> n mod m = 1 + (n - 1) mod m) proof (prove)
goal (1 subgoal):
1. 1 \<le> m \<Longrightarrow> (n mod m = 0 \<longrightarrow> m = 1 + (n - 1) mod m) \<and> (n mod m \<noteq> 0 \<longrightarrow> n mod m = 1 + (n - 1) mod m) proof (prove)
goal (2 subgoals):
1. \<lbrakk>1 \<le> m; m = 1\<rbrakk> \<Longrightarrow> (n mod m = 0 \<longrightarrow> m = 1 + (n - 1) mod m) \<and> (n mod m \<noteq> 0 \<longrightarrow> n mod m = 1 + (n - 1) mod m)
2. \<lbrakk>1 \<le> m; m \<noteq> 1\<rbrakk> \<Longrightarrow> (n mod m = 0 \<longrightarrow> m = 1 + (n - 1) mod m) \<and> (n mod m \<noteq> 0 \<longrightarrow> n mod m = 1 + (n - 1) mod m) proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1\<rbrakk> \<Longrightarrow> (n mod m = 0 \<longrightarrow> m = 1 + (n - 1) mod m) \<and> (n mod m \<noteq> 0 \<longrightarrow> n mod m = 1 + (n - 1) mod m) proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1\<rbrakk> \<Longrightarrow> n mod m \<noteq> 0 \<longrightarrow> n mod m = 1 + (n - 1) mod m proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> n mod m = 1 + (n - 1) mod m proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> (n - 1 + 1) mod m = 1 + (n - 1) mod m proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> ((n - 1) mod m + 1 mod m) mod m = 1 + (n - 1) mod m proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> (1 + (n - 1) mod m) mod m = 1 + (n - 1) mod m proof (prove)
goal (2 subgoals):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 0 \<le> 1 + (n - 1) mod m
2. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m proof (prove)
goal (2 subgoals):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 0 + 0 \<le> 1 + (n - 1) mod m
2. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m proof (prove)
goal (3 subgoals):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 0 \<le> 1
2. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 0 \<le> (n - 1) mod m
3. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m proof (prove)
goal (2 subgoals):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 0 \<le> (n - 1) mod m
2. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m proof (prove)
goal (2 subgoals):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1) mod m = m - 1\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m
2. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1) mod m \<noteq> m - 1\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m proof (prove)
goal (2 subgoals):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1 + 1) mod m = (m - 1 + 1) mod m\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m
2. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1) mod m \<noteq> m - 1\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1) mod m \<noteq> m - 1\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m proof (prove)
goal (2 subgoals):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1) mod m \<noteq> m - 1; 1 + (n - 1) mod m \<le> m\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m < m
2. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1) mod m \<noteq> m - 1\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m \<le> m proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1) mod m \<noteq> m - 1\<rbrakk> \<Longrightarrow> 1 + (n - 1) mod m \<le> m proof (prove)
goal (1 subgoal):
1. \<lbrakk>1 \<le> m; m \<noteq> 1; n mod m \<noteq> 0; (n - 1) mod m \<noteq> m - 1\<rbrakk> \<Longrightarrow> (n - 1) mod m < m proof (prove)
goal:
No subgoals! | lemma int_div_sub_1:
"\<lbrakk> m \<ge> 1 \<rbrakk> \<Longrightarrow> (n - (1 :: int)) div m = (if m dvd n then (n div m) - 1 else n div m)" | unnamed_thy_65 | More_Divides | 27 |
|
[] | lemma power_minus_is_div:
"b \<le> a \<Longrightarrow> (2 :: nat) ^ (a - b) = 2 ^ a div 2 ^ b" apply (induct a arbitrary: b) apply simp apply (erule le_SucE) apply (clarsimp simp:Suc_diff_le le_iff_add power_add) apply simp done | lemma power_minus_is_div:
"b \<le> a \<Longrightarrow> (2 :: nat) ^ (a - b) = 2 ^ a div 2 ^ b" apply (induct a arbitrary: b) apply simp apply (erule le_SucE) apply (clarsimp simp:Suc_diff_le le_iff_add power_add) apply simp done | proof (prove)
goal (1 subgoal):
1. b \<le> a \<Longrightarrow> 2 ^ (a - b) = 2 ^ a div 2 ^ b proof (prove)
goal (2 subgoals):
1. \<And>b. b \<le> 0 \<Longrightarrow> 2 ^ (0 - b) = 2 ^ 0 div 2 ^ b
2. \<And>a b. \<lbrakk>\<And>b. b \<le> a \<Longrightarrow> 2 ^ (a - b) = 2 ^ a div 2 ^ b; b \<le> Suc a\<rbrakk> \<Longrightarrow> 2 ^ (Suc a - b) = 2 ^ Suc a div 2 ^ b proof (prove)
goal (1 subgoal):
1. \<And>a b. \<lbrakk>\<And>b. b \<le> a \<Longrightarrow> 2 ^ (a - b) = 2 ^ a div 2 ^ b; b \<le> Suc a\<rbrakk> \<Longrightarrow> 2 ^ (Suc a - b) = 2 ^ Suc a div 2 ^ b proof (prove)
goal (2 subgoals):
1. \<And>a b. \<lbrakk>\<And>b. b \<le> a \<Longrightarrow> 2 ^ (a - b) = 2 ^ a div 2 ^ b; b \<le> a\<rbrakk> \<Longrightarrow> 2 ^ (Suc a - b) = 2 ^ Suc a div 2 ^ b
2. \<And>a b. \<lbrakk>\<And>b. b \<le> a \<Longrightarrow> 2 ^ (a - b) = 2 ^ a div 2 ^ b; b = Suc a\<rbrakk> \<Longrightarrow> 2 ^ (Suc a - b) = 2 ^ Suc a div 2 ^ b proof (prove)
goal (1 subgoal):
1. \<And>a b. \<lbrakk>\<And>b. b \<le> a \<Longrightarrow> 2 ^ (a - b) = 2 ^ a div 2 ^ b; b = Suc a\<rbrakk> \<Longrightarrow> 2 ^ (Suc a - b) = 2 ^ Suc a div 2 ^ b proof (prove)
goal:
No subgoals! | lemma power_minus_is_div:
"b \<le> a \<Longrightarrow> (2 :: nat) ^ (a - b) = 2 ^ a div 2 ^ b" | unnamed_thy_66 | More_Divides | 6 |
|
[] | lemma two_pow_div_gt_le:
"v < 2 ^ n div (2 ^ m :: nat) \<Longrightarrow> m \<le> n" by (clarsimp dest!: less_two_pow_divD) | lemma two_pow_div_gt_le:
"v < 2 ^ n div (2 ^ m :: nat) \<Longrightarrow> m \<le> n" by (clarsimp dest!: less_two_pow_divD) | proof (prove)
goal (1 subgoal):
1. v < 2 ^ n div 2 ^ m \<Longrightarrow> m \<le> n | lemma two_pow_div_gt_le:
"v < 2 ^ n div (2 ^ m :: nat) \<Longrightarrow> m \<le> n" | unnamed_thy_67 | More_Divides | 1 |
|
[] | lemma td_gal_lt:
\<open>0 < c \<Longrightarrow> a < b * c \<longleftrightarrow> a div c < b\<close>
for a b c :: nat by (simp add: div_less_iff_less_mult) | lemma td_gal_lt:
\<open>0 < c \<Longrightarrow> a < b * c \<longleftrightarrow> a div c < b\<close>
for a b c :: nat by (simp add: div_less_iff_less_mult) | proof (prove)
goal (1 subgoal):
1. 0 < c \<Longrightarrow> (a < b * c) = (a div c < b) | lemma td_gal_lt:
\<open>0 < c \<Longrightarrow> a < b * c \<longleftrightarrow> a div c < b\<close>
for a b c :: nat | unnamed_thy_68 | More_Divides | 1 |
|
[] | lemma td_gal:
\<open>0 < c \<Longrightarrow> b * c \<le> a \<longleftrightarrow> b \<le> a div c\<close>
for a b c :: nat by (simp add: less_eq_div_iff_mult_less_eq) | lemma td_gal:
\<open>0 < c \<Longrightarrow> b * c \<le> a \<longleftrightarrow> b \<le> a div c\<close>
for a b c :: nat by (simp add: less_eq_div_iff_mult_less_eq) | proof (prove)
goal (1 subgoal):
1. 0 < c \<Longrightarrow> (b * c \<le> a) = (b \<le> a div c) | lemma td_gal:
\<open>0 < c \<Longrightarrow> b * c \<le> a \<longleftrightarrow> b \<le> a div c\<close>
for a b c :: nat | unnamed_thy_69 | More_Divides | 1 |
|
[] | lemma disjunctive_add2:
"(x AND y) = 0 \<Longrightarrow> x + y = x OR y" by (metis disjunctive_add bit_0_eq bit_and_iff bot_apply bot_bool_def) | lemma disjunctive_add2:
"(x AND y) = 0 \<Longrightarrow> x + y = x OR y" by (metis disjunctive_add bit_0_eq bit_and_iff bot_apply bot_bool_def) | proof (prove)
goal (1 subgoal):
1. x AND y = (0::'a) \<Longrightarrow> x + y = x OR y | lemma disjunctive_add2:
"(x AND y) = 0 \<Longrightarrow> x + y = x OR y" | unnamed_thy_70 | More_Bit_Ring | 1 |
|
[] | lemma not_xor_is_eqv:
"NOT (x XOR y) = (x AND y) OR (NOT x AND NOT y)" by (simp add: bit.xor_def bit.disj_conj_distrib or.commute) | lemma not_xor_is_eqv:
"NOT (x XOR y) = (x AND y) OR (NOT x AND NOT y)" by (simp add: bit.xor_def bit.disj_conj_distrib or.commute) | proof (prove)
goal (1 subgoal):
1. NOT (x XOR y) = x AND y OR NOT x AND NOT y | lemma not_xor_is_eqv:
"NOT (x XOR y) = (x AND y) OR (NOT x AND NOT y)" | unnamed_thy_71 | More_Bit_Ring | 1 |
|
[] | lemma not_xor_eq_xor_not:
"(NOT x) XOR y = x XOR (NOT y)" by simp | lemma not_xor_eq_xor_not:
"(NOT x) XOR y = x XOR (NOT y)" by simp | proof (prove)
goal (1 subgoal):
1. NOT x XOR y = x XOR NOT y | lemma not_xor_eq_xor_not:
"(NOT x) XOR y = x XOR (NOT y)" | unnamed_thy_72 | More_Bit_Ring | 1 |
|
[] | lemma minus_not_eq_plus_1:
"- NOT x = x + 1" by (simp add: minus_eq_not_plus_1) | lemma minus_not_eq_plus_1:
"- NOT x = x + 1" by (simp add: minus_eq_not_plus_1) | proof (prove)
goal (1 subgoal):
1. - NOT x = x + (1::'a) | lemma minus_not_eq_plus_1:
"- NOT x = x + 1" | unnamed_thy_74 | More_Bit_Ring | 1 |
|
[] | lemma not_minus_eq_minus_1:
"NOT (- x) = x - 1" by (simp add: not_eq_complement) | lemma not_minus_eq_minus_1:
"NOT (- x) = x - 1" by (simp add: not_eq_complement) | proof (prove)
goal (1 subgoal):
1. NOT (- x) = x - (1::'a) | lemma not_minus_eq_minus_1:
"NOT (- x) = x - 1" | unnamed_thy_75 | More_Bit_Ring | 1 |
|
[] | lemma and_plus_not_and:
"(x AND y) + (x AND NOT y) = x" by (metis and.left_commute and.right_neutral bit.conj_cancel_right bit.conj_disj_distrib
bit.conj_zero_right bit.disj_cancel_right disjunctive_add2) | lemma and_plus_not_and:
"(x AND y) + (x AND NOT y) = x" by (metis and.left_commute and.right_neutral bit.conj_cancel_right bit.conj_disj_distrib
bit.conj_zero_right bit.disj_cancel_right disjunctive_add2) | proof (prove)
goal (1 subgoal):
1. (x AND y) + (x AND NOT y) = x | lemma and_plus_not_and:
"(x AND y) + (x AND NOT y) = x" | unnamed_thy_76 | More_Bit_Ring | 1 |
|
[] | lemma and_eq_not_or_minus:
"x AND y = (NOT x OR y) - NOT x" by (metis and.idem and_eq_not_not_or eq_diff_eq or.commute or.idem or_eq_and_not_plus) | lemma and_eq_not_or_minus:
"x AND y = (NOT x OR y) - NOT x" by (metis and.idem and_eq_not_not_or eq_diff_eq or.commute or.idem or_eq_and_not_plus) | proof (prove)
goal (1 subgoal):
1. x AND y = (NOT x OR y) - NOT x | lemma and_eq_not_or_minus:
"x AND y = (NOT x OR y) - NOT x" | unnamed_thy_78 | More_Bit_Ring | 1 |
|
[] | lemma and_not_eq_or_minus:
"x AND NOT y = (x OR y) - y" by (simp add: or_eq_and_not_plus) | lemma and_not_eq_or_minus:
"x AND NOT y = (x OR y) - y" by (simp add: or_eq_and_not_plus) | proof (prove)
goal (1 subgoal):
1. x AND NOT y = (x OR y) - y | lemma and_not_eq_or_minus:
"x AND NOT y = (x OR y) - y" | unnamed_thy_79 | More_Bit_Ring | 1 |
|
[] | lemma and_not_eq_minus_and:
"x AND NOT y = x - (x AND y)" by (simp add: add.commute eq_diff_eq and_plus_not_and) | lemma and_not_eq_minus_and:
"x AND NOT y = x - (x AND y)" by (simp add: add.commute eq_diff_eq and_plus_not_and) | proof (prove)
goal (1 subgoal):
1. x AND NOT y = x - (x AND y) | lemma and_not_eq_minus_and:
"x AND NOT y = x - (x AND y)" | unnamed_thy_80 | More_Bit_Ring | 1 |
|
[] | lemma or_minus_eq_minus_and:
"(x OR y) - y = x - (x AND y)" by (metis and_not_eq_minus_and and_not_eq_or_minus) | lemma or_minus_eq_minus_and:
"(x OR y) - y = x - (x AND y)" by (metis and_not_eq_minus_and and_not_eq_or_minus) | proof (prove)
goal (1 subgoal):
1. (x OR y) - y = x - (x AND y) | lemma or_minus_eq_minus_and:
"(x OR y) - y = x - (x AND y)" | unnamed_thy_81 | More_Bit_Ring | 1 |
|
[] | lemma plus_eq_and_or:
"x + y = (x OR y) + (x AND y)" using add_commute local.add.semigroup_axioms or_eq_and_not_plus semigroup.assoc by (fastforce simp: and_plus_not_and) | lemma plus_eq_and_or:
"x + y = (x OR y) + (x AND y)" using add_commute local.add.semigroup_axioms or_eq_and_not_plus semigroup.assoc by (fastforce simp: and_plus_not_and) | proof (prove)
goal (1 subgoal):
1. x + y = (x OR y) + (x AND y) proof (prove)
using this:
?a + ?b = ?b + ?a
semigroup (+)
?x OR ?y = (?x AND NOT ?y) + ?y
semigroup ?f \<Longrightarrow> ?f (?f ?a ?b) ?c = ?f ?a (?f ?b ?c)
goal (1 subgoal):
1. x + y = (x OR y) + (x AND y) | lemma plus_eq_and_or:
"x + y = (x OR y) + (x AND y)" | unnamed_thy_82 | More_Bit_Ring | 2 |
|
[] | lemma xor_eq_or_minus_and:
"x XOR y = (x OR y) - (x AND y)" by (metis (no_types) bit.de_Morgan_conj bit.xor_def2 bit_and_iff bit_or_iff disjunctive_diff) | lemma xor_eq_or_minus_and:
"x XOR y = (x OR y) - (x AND y)" by (metis (no_types) bit.de_Morgan_conj bit.xor_def2 bit_and_iff bit_or_iff disjunctive_diff) | proof (prove)
goal (1 subgoal):
1. x XOR y = (x OR y) - (x AND y) | lemma xor_eq_or_minus_and:
"x XOR y = (x OR y) - (x AND y)" | unnamed_thy_83 | More_Bit_Ring | 1 |
|
[] | lemma not_xor_eq_and_plus_not_or:
"NOT (x XOR y) = (x AND y) + NOT (x OR y)" by (metis (no_types, lifting) not_diff_distrib add.commute bit.de_Morgan_conj bit.xor_def2
bit_and_iff bit_or_iff disjunctive_diff) | lemma not_xor_eq_and_plus_not_or:
"NOT (x XOR y) = (x AND y) + NOT (x OR y)" by (metis (no_types, lifting) not_diff_distrib add.commute bit.de_Morgan_conj bit.xor_def2
bit_and_iff bit_or_iff disjunctive_diff) | proof (prove)
goal (1 subgoal):
1. NOT (x XOR y) = (x AND y) + NOT (x OR y) | lemma not_xor_eq_and_plus_not_or:
"NOT (x XOR y) = (x AND y) + NOT (x OR y)" | unnamed_thy_84 | More_Bit_Ring | 1 |
|
[] | lemma not_xor_eq_and_minus_or:
"NOT (x XOR y) = (x AND y) - (x OR y) - 1" by (metis not_diff_distrib add.commute minus_diff_eq not_minus_eq_minus_1 not_xor_eq_and_plus_not_or) | lemma not_xor_eq_and_minus_or:
"NOT (x XOR y) = (x AND y) - (x OR y) - 1" by (metis not_diff_distrib add.commute minus_diff_eq not_minus_eq_minus_1 not_xor_eq_and_plus_not_or) | proof (prove)
goal (1 subgoal):
1. NOT (x XOR y) = (x AND y) - (x OR y) - (1::'a) | lemma not_xor_eq_and_minus_or:
"NOT (x XOR y) = (x AND y) - (x OR y) - 1" | unnamed_thy_85 | More_Bit_Ring | 1 |
|
[] | lemma plus_eq_xor_plus_carry:
"x + y = (x XOR y) + 2 * (x AND y)" by (metis plus_eq_and_or add.commute add.left_commute diff_add_cancel mult_2 xor_eq_or_minus_and) | lemma plus_eq_xor_plus_carry:
"x + y = (x XOR y) + 2 * (x AND y)" by (metis plus_eq_and_or add.commute add.left_commute diff_add_cancel mult_2 xor_eq_or_minus_and) | proof (prove)
goal (1 subgoal):
1. x + y = (x XOR y) + (2::'a) * (x AND y) | lemma plus_eq_xor_plus_carry:
"x + y = (x XOR y) + 2 * (x AND y)" | unnamed_thy_86 | More_Bit_Ring | 1 |
|
[] | lemma plus_eq_or_minus_xor:
"x + y = 2 * (x OR y) - (x XOR y)" by (metis add_diff_cancel_left' diff_diff_eq2 local.mult_2 plus_eq_and_or xor_eq_or_minus_and) | lemma plus_eq_or_minus_xor:
"x + y = 2 * (x OR y) - (x XOR y)" by (metis add_diff_cancel_left' diff_diff_eq2 local.mult_2 plus_eq_and_or xor_eq_or_minus_and) | proof (prove)
goal (1 subgoal):
1. x + y = (2::'a) * (x OR y) - (x XOR y) | lemma plus_eq_or_minus_xor:
"x + y = 2 * (x OR y) - (x XOR y)" | unnamed_thy_87 | More_Bit_Ring | 1 |
|
[] | lemma plus_eq_minus_neg:
"x + y = x - NOT y - 1" using add_commute local.not_diff_distrib not_minus by auto | lemma plus_eq_minus_neg:
"x + y = x - NOT y - 1" using add_commute local.not_diff_distrib not_minus by auto | proof (prove)
goal (1 subgoal):
1. x + y = x - NOT y - (1::'a) proof (prove)
using this:
?a + ?b = ?b + ?a
NOT (?a - ?b) = NOT ?a + ?b
NOT (?x - ?y) = ?y - ?x - (1::'a)
goal (1 subgoal):
1. x + y = x - NOT y - (1::'a) | lemma plus_eq_minus_neg:
"x + y = x - NOT y - 1" | unnamed_thy_88 | More_Bit_Ring | 2 |
|
[] | lemma minus_eq_plus_neg:
"x - y = x + NOT y + 1" by (simp add: add.semigroup_axioms diff_conv_add_uminus minus_eq_not_plus_1 semigroup.assoc) | lemma minus_eq_plus_neg:
"x - y = x + NOT y + 1" by (simp add: add.semigroup_axioms diff_conv_add_uminus minus_eq_not_plus_1 semigroup.assoc) | proof (prove)
goal (1 subgoal):
1. x - y = x + NOT y + (1::'a) | lemma minus_eq_plus_neg:
"x - y = x + NOT y + 1" | unnamed_thy_89 | More_Bit_Ring | 1 |
|
[] | lemma minus_eq_and_not_minus_not_and:
"x - y = (x AND NOT y) - (NOT x AND y)" by (metis bit.de_Morgan_conj bit.double_compl not_diff_distrib plus_eq_and_or) | lemma minus_eq_and_not_minus_not_and:
"x - y = (x AND NOT y) - (NOT x AND y)" by (metis bit.de_Morgan_conj bit.double_compl not_diff_distrib plus_eq_and_or) | proof (prove)
goal (1 subgoal):
1. x - y = (x AND NOT y) - (NOT x AND y) | lemma minus_eq_and_not_minus_not_and:
"x - y = (x AND NOT y) - (NOT x AND y)" | unnamed_thy_90 | More_Bit_Ring | 1 |
|
[] | lemma minus_eq_xor_minus_not_and:
"x - y = (x XOR y) - 2 * (NOT x AND y)" by (metis (no_types) bit.compl_eq_compl_iff bit.xor_compl_left not_diff_distrib
plus_eq_xor_plus_carry) | lemma minus_eq_xor_minus_not_and:
"x - y = (x XOR y) - 2 * (NOT x AND y)" by (metis (no_types) bit.compl_eq_compl_iff bit.xor_compl_left not_diff_distrib
plus_eq_xor_plus_carry) | proof (prove)
goal (1 subgoal):
1. x - y = (x XOR y) - (2::'a) * (NOT x AND y) | lemma minus_eq_xor_minus_not_and:
"x - y = (x XOR y) - 2 * (NOT x AND y)" | unnamed_thy_91 | More_Bit_Ring | 1 |
|
[] | lemma minus_eq_and_not_minus_xor:
"x - y = 2 * (x AND NOT y) - (x XOR y)" by (metis and.commute minus_diff_eq minus_eq_xor_minus_not_and xor.commute) | lemma minus_eq_and_not_minus_xor:
"x - y = 2 * (x AND NOT y) - (x XOR y)" by (metis and.commute minus_diff_eq minus_eq_xor_minus_not_and xor.commute) | proof (prove)
goal (1 subgoal):
1. x - y = (2::'a) * (x AND NOT y) - (x XOR y) | lemma minus_eq_and_not_minus_xor:
"x - y = 2 * (x AND NOT y) - (x XOR y)" | unnamed_thy_92 | More_Bit_Ring | 1 |
|
[] | lemma and_one_neq_simps[simp]:
"x AND 1 \<noteq> 0 \<longleftrightarrow> x AND 1 = 1"
"x AND 1 \<noteq> 1 \<longleftrightarrow> x AND 1 = 0" by (clarsimp simp: and_one_eq)+ | lemma and_one_neq_simps[simp]:
"x AND 1 \<noteq> 0 \<longleftrightarrow> x AND 1 = 1"
"x AND 1 \<noteq> 1 \<longleftrightarrow> x AND 1 = 0" by (clarsimp simp: and_one_eq)+ | proof (prove)
goal (1 subgoal):
1. (x AND (1::'a) \<noteq> (0::'a)) = (x AND (1::'a) = (1::'a)) &&& (x AND (1::'a) \<noteq> (1::'a)) = (x AND (1::'a) = (0::'a)) | lemma and_one_neq_simps[simp]:
"x AND 1 \<noteq> 0 \<longleftrightarrow> x AND 1 = 1"
"x AND 1 \<noteq> 1 \<longleftrightarrow> x AND 1 = 0" | unnamed_thy_93 | More_Bit_Ring | 1 |
|
[] | lemma unat_power_lower [simp]:
"unat ((2::'a::len word) ^ n) = 2 ^ n" if "n < LENGTH('a::len)" using that by transfer simp | lemma unat_power_lower [simp]:
"unat ((2::'a::len word) ^ n) = 2 ^ n" if "n < LENGTH('a::len)" using that by transfer simp | proof (prove)
goal (1 subgoal):
1. unat (2 ^ n) = 2 ^ n proof (prove)
using this:
n < LENGTH('a)
goal (1 subgoal):
1. unat (2 ^ n) = 2 ^ n | lemma unat_power_lower [simp]:
"unat ((2::'a::len word) ^ n) = 2 ^ n" if "n < LENGTH('a::len)" | unnamed_thy_95 | More_Word | 2 |
|
[] | lemma word_div_lt_eq_0:
"x < y \<Longrightarrow> x div y = 0" for x :: "'a :: len word" by (fact div_word_less) | lemma word_div_lt_eq_0:
"x < y \<Longrightarrow> x div y = 0" for x :: "'a :: len word" by (fact div_word_less) | proof (prove)
goal (1 subgoal):
1. x < y \<Longrightarrow> x div y = 0 | lemma word_div_lt_eq_0:
"x < y \<Longrightarrow> x div y = 0" for x :: "'a :: len word" | unnamed_thy_97 | More_Word | 1 |
|
[] | lemma word_div_eq_1_iff: "n div m = 1 \<longleftrightarrow> n \<ge> m \<and> unat n < 2 * unat (m :: 'a :: len word)" apply (simp only: word_arith_nat_defs word_le_nat_alt word_of_nat_eq_iff flip: nat_div_eq_Suc_0_iff) apply (simp flip: unat_div unsigned_take_bit_eq) done | lemma word_div_eq_1_iff: "n div m = 1 \<longleftrightarrow> n \<ge> m \<and> unat n < 2 * unat (m :: 'a :: len word)" apply (simp only: word_arith_nat_defs word_le_nat_alt word_of_nat_eq_iff flip: nat_div_eq_Suc_0_iff) apply (simp flip: unat_div unsigned_take_bit_eq) done | proof (prove)
goal (1 subgoal):
1. (n div m = 1) = (m \<le> n \<and> unat n < 2 * unat m) proof (prove)
goal (1 subgoal):
1. (take_bit LENGTH('a) (unat n div unat m) = take_bit LENGTH('a) (Suc 0)) = (unat n div unat m = Suc 0) proof (prove)
goal:
No subgoals! | lemma word_div_eq_1_iff: "n div m = 1 \<longleftrightarrow> n \<ge> m \<and> unat n < 2 * unat (m :: 'a :: len word)" | unnamed_thy_98 | More_Word | 3 |
|
[] | lemma AND_twice [simp]:
"(w AND m) AND m = w AND m" by (fact and.right_idem) | lemma AND_twice [simp]:
"(w AND m) AND m = w AND m" by (fact and.right_idem) | proof (prove)
goal (1 subgoal):
1. (w AND m) AND m = w AND m | lemma AND_twice [simp]:
"(w AND m) AND m = w AND m" | unnamed_thy_99 | More_Word | 1 |
|
[] | lemma word_combine_masks:
"w AND m = z \<Longrightarrow> w AND m' = z' \<Longrightarrow> w AND (m OR m') = (z OR z')"
for w m m' z z' :: \<open>'a::len word\<close> by (simp add: bit.conj_disj_distrib) | lemma word_combine_masks:
"w AND m = z \<Longrightarrow> w AND m' = z' \<Longrightarrow> w AND (m OR m') = (z OR z')"
for w m m' z z' :: \<open>'a::len word\<close> by (simp add: bit.conj_disj_distrib) | proof (prove)
goal (1 subgoal):
1. \<lbrakk>w AND m = z; w AND m' = z'\<rbrakk> \<Longrightarrow> w AND (m OR m') = z OR z' | lemma word_combine_masks:
"w AND m = z \<Longrightarrow> w AND m' = z' \<Longrightarrow> w AND (m OR m') = (z OR z')"
for w m m' z z' :: \<open>'a::len word\<close> | unnamed_thy_100 | More_Word | 1 |
|
[] | lemma p2_gt_0:
"(0 < (2 ^ n :: 'a :: len word)) = (n < LENGTH('a))" by (simp add : word_gt_0 not_le) | lemma p2_gt_0:
"(0 < (2 ^ n :: 'a :: len word)) = (n < LENGTH('a))" by (simp add : word_gt_0 not_le) | proof (prove)
goal (1 subgoal):
1. (0 < 2 ^ n) = (n < LENGTH('a)) | lemma p2_gt_0:
"(0 < (2 ^ n :: 'a :: len word)) = (n < LENGTH('a))" | unnamed_thy_101 | More_Word | 1 |
|
[] | lemma uint_2p_alt:
\<open>n < LENGTH('a::len) \<Longrightarrow> uint ((2::'a::len word) ^ n) = 2 ^ n\<close> using p2_gt_0 [of n, where ?'a = 'a] by (simp add: uint_2p) | lemma uint_2p_alt:
\<open>n < LENGTH('a::len) \<Longrightarrow> uint ((2::'a::len word) ^ n) = 2 ^ n\<close> using p2_gt_0 [of n, where ?'a = 'a] by (simp add: uint_2p) | proof (prove)
goal (1 subgoal):
1. n < LENGTH('a) \<Longrightarrow> uint (2 ^ n) = 2 ^ n proof (prove)
using this:
(0 < 2 ^ n) = (n < LENGTH('a))
goal (1 subgoal):
1. n < LENGTH('a) \<Longrightarrow> uint (2 ^ n) = 2 ^ n | lemma uint_2p_alt:
\<open>n < LENGTH('a::len) \<Longrightarrow> uint ((2::'a::len word) ^ n) = 2 ^ n\<close> | unnamed_thy_102 | More_Word | 2 |
|
[] | lemma p2_eq_0:
\<open>(2::'a::len word) ^ n = 0 \<longleftrightarrow> LENGTH('a::len) \<le> n\<close> by (fact exp_eq_zero_iff) | lemma p2_eq_0:
\<open>(2::'a::len word) ^ n = 0 \<longleftrightarrow> LENGTH('a::len) \<le> n\<close> by (fact exp_eq_zero_iff) | proof (prove)
goal (1 subgoal):
1. (2 ^ n = 0) = (LENGTH('a) \<le> n) | lemma p2_eq_0:
\<open>(2::'a::len word) ^ n = 0 \<longleftrightarrow> LENGTH('a::len) \<le> n\<close> | unnamed_thy_103 | More_Word | 1 |
|
[] | lemma p2len:
\<open>(2 :: 'a word) ^ LENGTH('a::len) = 0\<close> by (fact word_pow_0) | lemma p2len:
\<open>(2 :: 'a word) ^ LENGTH('a::len) = 0\<close> by (fact word_pow_0) | proof (prove)
goal (1 subgoal):
1. 2 ^ LENGTH('a) = 0 | lemma p2len:
\<open>(2 :: 'a word) ^ LENGTH('a::len) = 0\<close> | unnamed_thy_104 | More_Word | 1 |
|
[] | lemma neg_mask_is_div:
"w AND NOT (mask n) = (w div 2^n) * 2^n"
for w :: \<open>'a::len word\<close> by (rule bit_word_eqI)
(auto simp add: bit_simps simp flip: push_bit_eq_mult drop_bit_eq_div) | lemma neg_mask_is_div:
"w AND NOT (mask n) = (w div 2^n) * 2^n"
for w :: \<open>'a::len word\<close> by (rule bit_word_eqI)
(auto simp add: bit_simps simp flip: push_bit_eq_mult drop_bit_eq_div) | proof (prove)
goal (1 subgoal):
1. w AND NOT (mask n) = w div 2 ^ n * 2 ^ n | lemma neg_mask_is_div:
"w AND NOT (mask n) = (w div 2^n) * 2^n"
for w :: \<open>'a::len word\<close> | unnamed_thy_105 | More_Word | 1 |
|
[] | lemma neg_mask_is_div':
"n < size w \<Longrightarrow> w AND NOT (mask n) = ((w div (2 ^ n)) * (2 ^ n))"
for w :: \<open>'a::len word\<close> by (rule neg_mask_is_div) | lemma neg_mask_is_div':
"n < size w \<Longrightarrow> w AND NOT (mask n) = ((w div (2 ^ n)) * (2 ^ n))"
for w :: \<open>'a::len word\<close> by (rule neg_mask_is_div) | proof (prove)
goal (1 subgoal):
1. n < size w \<Longrightarrow> w AND NOT (mask n) = w div 2 ^ n * 2 ^ n | lemma neg_mask_is_div':
"n < size w \<Longrightarrow> w AND NOT (mask n) = ((w div (2 ^ n)) * (2 ^ n))"
for w :: \<open>'a::len word\<close> | unnamed_thy_106 | More_Word | 1 |
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