IMAP UID proof
Notations
 $h$: the hash of a message, $\mathbb{H}$ is the set of hashes
 $i$: the UID of a message $(i \in \mathbb{N})$
 $f$: a flag attributed to a message (it's a string), we write $\mathbb{F}$ the set of possible flags
 if $M$ is a map (aka a dictionnary), if $x$ has no assigned value in $M$ we write $M [x] = \bot$ or equivalently $x \not\in M$. If $x$ has a value in the map we write $x \in M$ and $M [x] \neq \bot$
State

A map $I$ such that $I [h]$ is the UID of the message whose hash is $h$ is the mailbox, or $\bot$ if there is no such message

A map $F$ such that $F [h]$ is the set of flags attributed to the message whose hash is $h$

$v$: the UIDVALIDITY value

$n$: the UIDNEXT value

$s$: an internal sequence number that is mostly equal to UIDNEXT but also grows when mails are deleted
Operations

MAIL_ADD$(h, i)$: the value of $i$ that is put in this operation is the value of $s$ in the state resulting of all already known operations, i.e. $s (O_{gen})$ in the notation below where $O_{gen}$ is the set of all operations known at the time when the MAIL_ADD is generated. Moreover, such an operation can only be generated if $I (O_{gen}) [h] = \bot$, i.e. for a mail $h$ that is not already in the state at $O_{gen}$.

MAIL_DEL$(h)$

FLAG_ADD$(h, f)$

FLAG_DEL$(h, f)$
Algorithms
apply MAIL_ADD$(h, i)$:
if $i < s$:
$v \leftarrow v + s  i$
if $F [h] = \bot$:
$F [h] \leftarrow F_{initial}$
$I [h] \leftarrow s$
$s \leftarrow s + 1$
$n \leftarrow s$
apply MAIL_DEL$(h)$:
$I [h] \leftarrow \bot$
$F [h] \leftarrow \bot$
$s \leftarrow s + 1$
apply FLAG_ADD$(h, f)$:
if $h \in F$:
$F [h] \leftarrow F [h] \cup { f }$
apply FLAG_DEL$(h, f)$:
if $h \in F$:
$F [h] \leftarrow F [h] \backslash { f }$
More notations

$o$ is an operation such as MAIL_ADD, MAIL_DEL, etc. $O$ is a set of operations. Operations embed a timestamp, so a set of operations $O$ can be written as $O = [o_1, o_2, \ldots, o_n]$ by ordering them by timestamp.

if $o \in O$, we write $O_{\leqslant o}$, $O_{< o}$, $O_{\geqslant o}$, $O_{> o}$ the set of items of $O$ that are respectively earlier or equal, strictly earlier, later or equal, or strictly later than $o$. In other words, if we write $O = [o_1, \ldots, o_n]$, where $o$ is a certain $o_i$ in this sequence, then: $$ \begin{aligned} O_{\leqslant o} &= { o_1, \ldots, o_i }\ O_{< o} &= { o_1, \ldots, o_{i  1} }\ O_{\geqslant o} &= { o_i, \ldots, o_n }\ O_{> o} &= { o_{i + 1}, \ldots, o_n } \end{aligned} $$

If $O$ is a set of operations, we write $I (O)$, $F (O)$, $n (O), s (O)$, and $v (O)$ the values of $I, F, n, s$ and $v$ in the state that results of applying all of the operations in $O$ in their sorted order. (we thus write $I (O) [h]$ the value of $I [h]$ in this state)
Hypothesis: An operation $o$ can only be in a set $O$ if it was generated after applying operations of a set $O_{gen}$ such that $O_{gen} \subset O$ (because causality is respected in how we deliver operations). Sets of operations that do not respect this property are excluded from all of the properties, lemmas and proofs below.
Simplification: We will now exclude FLAG_ADD and FLAG_DEL operations, as they do not manipulate $n$, $s$ and $v$, and adding them should have no impact on the properties below.
Small lemma: If there are no FLAG_ADD and FLAG_DEL operations, then $s (O) =  O $. This is easy to see because the possible operations are only MAIL_ADD and MAIL_DEL, and both increment the value of $s$ by 1.
Defnition: If $o$ is a MAIL_ADD$(h, i)$ operation, and $O$ is a set of operations such that $o \in O$, then we define the following value: $$ C (o, O) = s (O_{< o})  i $$ We say that $C (o, O)$ is the number of conflicts of $o$ in $O$: it corresponds to the number of operations that were added before $o$ in $O$ that were not in $O_{gen}$.
Property:
We have that:
$$ v (O) = \sum_{o \in O} C (o, O) $$
Or in English: $v (O)$ is the sum of the number of conflicts of all of the MAIL_ADD operations in $O$. This is easy to see because indeed $v$ is incremented by $C (o, O)$ for each operation $o \in O$ that is applied.
Property: If $O$ and $O'$ are two sets of operations, and $O \subseteq O'$, then:
$$ \begin{aligned} \forall o \in O, \qquad C (o, O) \leqslant C (o, O') \end{aligned} $$
This is easy to see because $O_{< o} \subseteq O'{< o}$ and $C (o, O')  C (o, O) = s (O'{< o})  s (O_{< o}) =  O'{< o}    O{< o}  \geqslant 0$
Theorem:
If $O$ and $O'$ are two sets of operations:
$$ \begin{aligned} O \subseteq O' & \Rightarrow & v (O) \leqslant v (O') \end{aligned} $$
Proof:
$$ \begin{aligned} v (O') &= \sum_{o \in O'} C (o, O')\ & \geqslant \sum_{o \in O} C (o, O') \qquad \text{(because $O \subseteq O'$)}\ & \geqslant \sum_{o \in O} C (o, O) \qquad \text{(because $\forall o \in O, C (o, O) \leqslant C (o, O')$)}\ & \geqslant v (O) \end{aligned} $$
Theorem:
If $O$ and $O'$ are two sets of operations, such that $O \subset O'$,
and if there are two different mails $h$ and $h'$ $(h \neq h')$ such that $I (O) [h] = I (O') [h']$
then: $$v (O) < v (O')$$
Proof:
We already know that $v (O) \leqslant v (O')$ because of the previous theorem. We will now look at the sum: $$ v (O') = \sum_{o \in O'} C (o, O') $$ and show that there is at least one term in this sum that is strictly larger than the corresponding term in the other sum: $$ v (O) = \sum_{o \in O} C (o, O) $$ Let $o$ be the last MAIL_ADD$(h, _)$ operation in $O$, i.e. the operation that gives its definitive UID to mail $h$ in $O$, and similarly $o'$ be the last MAIL_ADD($h', _$) operation in $O'$.
Let us write $I = I (O) [h] = I (O') [h']$
$o$ is the operation at position $I$ in $O$, and $o'$ is the operation at position $I$ in $O'$. But $o \neq o'$, so if $o$ is not the operation at position $I$ in $O'$ then it has to be at a later position $I' > I$ in $O'$, because no operations are removed between $O$ and $O'$, the only possibility is that some other operations (including $o'$) are added before $o$. Therefore we have that $C (o, O') > C (o, O)$, i.e. at least one term in the sum above is strictly larger in the first sum than in the second one. Since all other terms are greater or equal, we have $v (O') > v (O)$.