# [isabelle] Lemma: Library/RBT.map-of-alist-of

```Hi all,

in the current distribution, the lemma "Library/RBT.map-of-alist-of" is
not shown, but marked with an "oops".
The whole development of folding an RB-tree is marked with "text {* The
following is still incomplete... *}".

In case no one has already proven this lemma, and someone is interested
in a proof, here is one.

Regards,
Peter
```
```theory RBT_add
imports RBT
begin

(* The next two lemmas are in my standard simpset, however I made them explicit for this proof: *)
"\<lbrakk> m\<in>dom l2 \<rbrakk> \<Longrightarrow> (l1++l2) m = l2 m"
"\<lbrakk> m\<notin>dom l1 \<rbrakk> \<Longrightarrow> (l1++l2) m = l2 m"
"\<lbrakk> m\<notin>dom l2 \<rbrakk> \<Longrightarrow> (l1++l2) m = l1 m"

lemma map_add_upd2: "m\<notin>dom e2 \<Longrightarrow> e1(m \<mapsto> u1) ++ e2 = (e1 ++ e2)(m \<mapsto> u1)"
apply (rule ext)
apply (auto split: option.split)
done

lemma map_of_alist_of_aux: "st (Tr c t1 k v t2) \<Longrightarrow> RBT.map_of (Tr c t1 k v t2) = RBT.map_of t2 ++ [k\<mapsto>v] ++ RBT.map_of t1"
proof (rule ext)
fix x
assume ST: "st (Tr c t1 k v t2)"
let ?thesis = "RBT.map_of (Tr c t1 k v t2) x = (RBT.map_of t2 ++ [k \<mapsto> v] ++ RBT.map_of t1) x"

have DOM_T1: "!!k'. k'\<in>dom (RBT.map_of t1) \<Longrightarrow> k>k'"
proof -
fix k'
from ST have "t1 |\<guillemotleft> k" by simp
with tlt_prop have "\<forall>k'\<in>keys t1. k>k'" by auto
moreover assume "k'\<in>dom (RBT.map_of t1)"
ultimately show "k>k'" using RBT.mapof_keys ST by auto
qed

have DOM_T2: "!!k'. k'\<in>dom (RBT.map_of t2) \<Longrightarrow> k<k'"
proof -
fix k'
from ST have "k \<guillemotleft>| t2" by simp
with tgt_prop have "\<forall>k'\<in>keys t2. k<k'" by auto
moreover assume "k'\<in>dom (RBT.map_of t2)"
ultimately show "k<k'" using RBT.mapof_keys ST by auto
qed

{
assume C: "x<k"
hence "RBT.map_of (Tr c t1 k v t2) x = RBT.map_of t1 x" by simp
moreover from C have "x\<notin>dom [k\<mapsto>v]" by simp
moreover have "x\<notin>dom (RBT.map_of t2)" proof
assume "x\<in>dom (RBT.map_of t2)"
with DOM_T2 have "k<x" by blast
with C show False by simp
qed
} moreover {
assume [simp]: "x=k"
hence "RBT.map_of (Tr c t1 k v t2) x = [k \<mapsto> v] x" by simp
moreover have "x\<notin>dom (RBT.map_of t1)" proof
assume "x\<in>dom (RBT.map_of t1)"
with DOM_T1 have "k>x" by blast
thus False by simp
qed
} moreover {
assume C: "x>k"
hence "RBT.map_of (Tr c t1 k v t2) x = RBT.map_of t2 x" by (simp add: less_not_sym[of k x])
moreover from C have "x\<notin>dom [k\<mapsto>v]" by simp
moreover have "x\<notin>dom (RBT.map_of t1)" proof
assume "x\<in>dom (RBT.map_of t1)"
with DOM_T1 have "k>x" by simp
with C show False by simp
qed
} ultimately show ?thesis using less_linear by blast
qed

(* This one is marked with an oops in RBT.thy *)
lemma map_of_alist_of:
shows "st t \<Longrightarrow> Map.map_of (alist_of t) = map_of t"
proof (induct t)
case Empty thus ?case by (simp add: RBT.map_of_Empty)
next
case (Tr c t1 k v t2)
hence "Map.map_of (alist_of (Tr c t1 k v t2)) = RBT.map_of t2 ++ [k \<mapsto> v] ++ RBT.map_of t1" by simp
also note map_of_alist_of_aux[OF Tr.prems,symmetric]
finally show ?case .
qed

end
```

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