# Re: [isabelle] remdups_adj crucial lemma

```On 26.08.2014 12:07, René Thiemann wrote:
> And since I already proved soundness (also in a lengthy version), I copied my proof into Lars
> file, which now contains the equivalence proof.
A little bit of cleanup, but no significant shortening.

```
```theory Scratch imports Main begin

lemma mono_image_least:
assumes f_mono: "mono f" and f_img: "f ` {m ..< n} = {m' ..< n'}" "m < n"
shows "f m = m'"
proof -
from f_img have "{m' ..< n'} \<noteq> {}"
by (metis atLeastLessThan_empty_iff image_is_empty)
with f_img have "m' \<in> f ` {m ..< n}" by auto
then obtain k where "f k = m'" "m \<le> k" by auto
moreover have "m' \<le> f m" using f_img by auto
ultimately show "f m = m'"
using f_mono by (auto elim: monoE[where x=m and y=k])
qed

lemma complete: "(\<exists>f :: nat \<Rightarrow> nat. mono f \<and> f ` {0 ..< size xs} = {0 ..< size ys}
\<and> (\<forall>i < size xs. xs!i = ys! (f i))
\<and> (\<forall>i. i + 1 < size xs \<longrightarrow> (xs!i = xs!(i+1) \<longleftrightarrow> f i = f(i+1))))
proof (induct xs arbitrary: ys rule: remdups_adj.induct)
case 1 then show ?case by auto
next
case (2 x) then obtain f where f_img: "f ` {0 ..< size [x]} = {0 ..< size ys}"
and f_nth: "\<And>i. i < size [x] \<Longrightarrow> [x]!i = ys!(f i)"
by metis

have "length ys = card (f ` {0 ..< size [x]})"
using f_img by auto
then have "length ys = 1" by auto
moreover
then have "f 0 = 0" using f_img by auto
ultimately show ?case using f_nth by (cases ys) auto
next
case (3 x1 x2 xs)
from "3.prems" obtain f where f_mono: "mono f"
and f_img: "f ` {0..<length (x1 # x2 # xs)} = {0..<length ys}"
and f_nth1: "\<And>i. i < length (x1 # x2 # xs) \<Longrightarrow> (x1 # x2 # xs) ! i = ys ! f i"
and f_nth2: "\<And>i. i + 1 < length (x1 # x2 #xs) \<Longrightarrow>
((x1 # x2 # xs) ! i = (x1 # x2 # xs) ! (i + 1)) = (f i = f (i + 1))"
by metis

show ?case
proof cases
assume "x1 = x2"

let ?f' = "f o Suc"

have "remdups_adj (x1 # xs) = ys"
proof (intro "3.hyps" exI conjI impI allI)
show "x1 = x2" by fact
next
show "mono ?f'"
using f_mono by (simp add: mono_iff_le_Suc)
next
have "?f' ` {0 ..< length (x1 # xs)} = f ` {Suc 0 ..< length (x1 # x2 # xs)}"
by safe (fastforce, rename_tac x, case_tac x, auto)
also have "\<dots> = f ` {0 ..< length (x1 # x2 # xs)}"
proof -
have "f 0 = f (Suc 0)" using \<open>x1 = x2\<close> f_nth2[of 0] by simp
then show ?thesis by safe (fastforce, rename_tac x, case_tac x, auto)
qed
also have "\<dots> = {0 ..< length ys}" by fact
finally show  "?f' ` {0 ..< length (x1 # xs)} = {0 ..< length ys}" .
next
fix i assume "i < length (x1 # xs)"
then show "(x1 # xs) ! i = ys ! ?f' i"
using f_nth1[of "Suc i"] \<open>x1 = x2\<close> by simp
next
fix i assume "i + 1 < length (x1 # xs)"
then show "((x1 # xs) ! i = (x1 # xs) ! (i + 1)) = (?f' i = ?f' (i + 1))"
using f_nth2[of "Suc i"] \<open>x1 = x2\<close> by simp_all
qed
then show ?thesis using \<open>x1 = x2\<close> by simp
next
assume "x1 \<noteq> x2"

then have "f 0 \<noteq> f (Suc 0)" using f_nth2[of 0] by auto

have "2 \<le> length ys"
proof -
have "2 = card {f 0, f 1}" using \<open>f 0 \<noteq> _\<close> by simp
also have "\<dots> \<le> card (f ` ({0, 1} \<union> {2..< Suc (Suc (length xs))}))" (is "_ \<le> card (f ` ?S)")
by (rule card_mono) auto
also have "?S = {0..<Suc (Suc (length xs))}" by auto
also have "card (f ` \<dots>) = length ys" using f_img by simp
finally show "2 \<le> length ys" .
qed

have "f 0 = 0" using f_mono f_img by (rule mono_image_least) simp

have "f (Suc 0) = Suc 0"
proof (rule ccontr)
assume "f (Suc 0) \<noteq> Suc 0"
then have "Suc 0 < f (Suc 0)" using f_nth2[of 0] \<open>x1 \<noteq> x2\<close> \<open>f 0 = 0\<close> by auto
then have "\<And>i. Suc 0 < f (Suc i)"
using f_mono by (metis Suc_le_mono le0 less_le_trans monoD)
then have "\<And>i. Suc 0 \<noteq> f i" using \<open>f 0 = 0\<close>
by (metis less_irrefl_nat not0_implies_Suc)
then have "Suc 0 \<notin> f ` {0 ..< length (x1 # x2 # xs)}" by auto
then show False using f_img \<open>2 \<le> length ys\<close> by auto
qed

obtain ys' where "ys = x1 # x2 # ys'"
using \<open>2 \<le> length ys\<close> f_nth1[of 0] f_nth1[of 1]
apply (cases ys)
apply (rename_tac  ys', case_tac  ys')
by (auto simp: \<open>f 0 = 0\<close> \<open>f (Suc 0) = Suc 0\<close>)

def f' \<equiv> "\<lambda>x. f (Suc x) - 1"

{ fix i
have "Suc 0 \<le> f (Suc 0)" using f_nth2[of 0] \<open>x1 \<noteq> x2\<close> \<open>f 0 = 0\<close>  by auto
also have "\<dots> \<le> f (Suc i)" using f_mono by (rule monoD) arith
finally have "Suc 0 \<le> f (Suc i)" . }
note Suc0_le_f_Suc = this

{ fix i have "f (Suc i) = Suc (f' i)"
using Suc0_le_f_Suc[of i] by (auto simp: f'_def) }
note f_Suc = this

have "remdups_adj (x2 # xs) = (x2 # ys')"
proof (intro "3.hyps" exI conjI impI allI)
show "x1 \<noteq> x2" by fact
next
show "mono f'" using Suc0_le_f_Suc f_mono by (auto simp: f'_def mono_iff_le_Suc le_diff_iff)
next
have "f' ` {0 ..< length (x2 # xs)} = (\<lambda>x. f x - 1) ` {0 ..< length (x1 # x2 #xs)}"
apply safe
apply (rename_tac [!] n,  case_tac [!] n)
apply (auto simp: f'_def \<open>f 0 = 0\<close> \<open>f (Suc 0) = Suc 0\<close> intro: rev_image_eqI)
done
also have "\<dots> = (\<lambda>x. x - 1) ` f ` {0 ..< length (x1 # x2 #xs)}"
by (auto simp: image_comp)
also have "\<dots> = (\<lambda>x. x - 1) ` {0 ..< length ys}"
by (simp only: f_img)
also have "\<dots> = {0 ..< length (x2 # ys')}"
using \<open>ys = _\<close> by (fastforce intro: rev_image_eqI)
finally show  "f' ` {0 ..< length (x2 # xs)} = {0 ..< length (x2 # ys')}" .
next
fix i assume  "i < length (x2 # xs)"
then show "(x2 # xs) ! i = (x2 # ys') ! f' i"
using f_nth1[of "Suc i"] \<open>x1 \<noteq> x2\<close> f_Suc by (simp add: \<open>ys = _\<close>)
next
fix i assume "i + 1 < length (x2 # xs)"
then show "((x2 # xs) ! i = (x2 # xs) ! (i + 1)) = (f' i = f' (i + 1))"
using f_nth2[of "Suc i"] \<open>x1 \<noteq> x2\<close> f_Suc by (auto simp: \<open>ys = _\<close>)
qed
then show ?case using \<open>ys = _\<close> \<open>x1 \<noteq> x2\<close> by simp
qed
qed

lemma sound: assumes "remdups_adj xs = ys"
shows "\<exists> f::nat => nat. mono f & f ` {0 ..< size xs} = {0 ..< size ys}
\<and> (\<forall> i < size xs. xs!i = ys!(f i))
\<and> (\<forall> i. i + 1 < size xs \<longrightarrow> (xs!i = xs!(i+1) <-> f i = f(i+1)))" (is "\<exists> f. ?p f xs ys")
using assms
proof (induct xs arbitrary: ys rule: remdups_adj.induct)
case (1 ys)
thus ?case
by (intro exI[of _ id], auto simp: mono_def)
next
case (2 x ys)
hence ys: "ys = [x]" by auto
show ?case unfolding ys
by (intro exI[of _ id], auto simp: mono_def)
next
case (3 x1 x2 xs ys)
let ?xs = "x1 # x2 # xs"
def zs \<equiv> "remdups_adj (x2 # xs)"
from 3(1-2)[of zs]
obtain f where p: "?p f (x2 # xs) zs" unfolding zs_def by (cases "x1 = x2") auto
then have f0: "f 0 = 0"
by (intro mono_image_least[where f=f]) blast+
from p have mono: "mono f" and f_xs_zs: "f ` {0..<length (x2 # xs)} = {0..<length zs}" by auto
have ys: "ys = (if x1 = x2 then zs else x1 # zs)"
unfolding 3(3)[symmetric] zs_def by auto
have zs0: "zs ! 0 = x2" unfolding zs_def by (induct xs) auto
{
fix i and g :: "nat \<Rightarrow> nat"
assume "i < Suc (length xs)"
hence "Suc i \<in> {0..<Suc (Suc (length xs))} \<inter> Collect (op < 0)" by auto
from imageI[OF this, of "(\<lambda>i. g (f (i - Suc 0)))"]
have "g (f i) \<in> (\<lambda>i. g (f (i - Suc 0))) ` ({0..<Suc (Suc (length xs))} \<inter> Collect (op < 0))" by auto
} note tedious = this
show ?case
proof (cases "x1 = x2")
case True
with ys have ys: "ys = zs" by auto
let ?f = "\<lambda> i. if i = 0 then 0 else f (i - 1)"
show ?thesis unfolding ys
proof (intro exI[of _ ?f] conjI allI impI)
fix i
assume i: "i < length ?xs"
with p show "?xs ! i = zs ! (?f i)" by (auto simp: zs0 True)
next
fix i
assume i: "i + 1 < length ?xs"
with p show "(?xs ! i = ?xs ! (i + 1)) = (?f i = ?f (i + 1))"
by (cases i) (auto simp: True f0 )
next
show "mono ?f" using f0 `mono f`
unfolding mono_def by auto
next
show "?f ` {0 ..< length ?xs} = {0 ..< length zs}" unfolding f_xs_zs[symmetric] using f0
tedious[of _ id] by auto
qed
next
case False
with ys have ys: "ys = x1 # zs" by auto
let ?f = "\<lambda> i. if i = 0 then 0 else Suc (f (i - 1))"
show ?thesis unfolding ys
proof (intro exI[of _ ?f] conjI allI impI)
fix i
assume i: "i < length ?xs"
with p show "?xs ! i = (x1 # zs) ! (?f i)" by auto
next
fix i
assume i: "i + 1 < length ?xs"
with p show "(?xs ! i = ?xs ! (i + 1)) = (?f i = ?f (i + 1))"
by (cases i) (auto simp: False f0 )
next
show "mono ?f" using f0 `mono f`
unfolding mono_def by auto
next
have id: "{0 ..< length (x1 # zs)} = insert 0 (Suc ` {0 ..< length zs})" by auto
show "?f ` {0 ..< length ?xs} = {0 ..< length (x1 # zs)}"
unfolding id f_xs_zs[symmetric] using tedious[of _ Suc]
by auto
qed
qed
qed

lemma equivalent: "(remdups_adj xs = ys) \<longleftrightarrow>
(\<exists> f::nat => nat. mono f & f ` {0 ..< size xs} = {0 ..< size ys}
\<and> (\<forall> i < size xs. xs!i = ys!(f i))
\<and> (\<forall> i. i + 1 < size xs \<longrightarrow> (xs!i = xs!(i+1) <-> f i = f(i+1))))"
using sound complete ..

end

```

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