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+# 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$(h, i)$:  
+&nbsp;&nbsp; *if* $i < s$:  
+&nbsp;&nbsp;&nbsp;&nbsp; $v \leftarrow v + s - i$  
+&nbsp;&nbsp; *if* $F [h] = \bot$:  
+&nbsp;&nbsp;&nbsp;&nbsp; $F [h] \leftarrow F_{initial}$  
+&nbsp;&nbsp;$I [h] \leftarrow s$  
+&nbsp;&nbsp;$s \leftarrow s + 1$  
+&nbsp;&nbsp;$n \leftarrow s$  
+
+**apply** MAIL\_DEL$(h)$:  
+&nbsp;&nbsp; $I [h] \leftarrow \bot$  
+&nbsp;&nbsp;$F [h] \leftarrow \bot$  
+&nbsp;&nbsp;$s \leftarrow s + 1$
+
+**apply** FLAG\_ADD$(h, f)$:  
+&nbsp;&nbsp; *if* $h \in F$:  
+&nbsp;&nbsp;&nbsp;&nbsp; $F [h] \leftarrow F [h] \cup \{ f \}$  
+
+**apply** FLAG\_DEL$(h, f)$:  
+&nbsp;&nbsp; *if* $h \in F$:  
+&nbsp;&nbsp;&nbsp;&nbsp; $F [h] \leftarrow F [h] \backslash \{ f \}$  
+
+
+**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 $v (O') > v (O)$.