**Base of the Canonical Extension : $\beta_0$**

*Definition 1:*If a sequence of integers $C$ does not possess the property of normality (see Definition 1, here), then there exists $c,c'$ in the interval $(0,n)$ such that $c\notin C,c'\notin C$, $c + c' - n = a + b$ where $C = A+B$ and $a\in A$ and $b\in B$

Let $\beta_0\in B$ be the least number $b$ for which $c + c' - n = a + b$

**Canonical Extension of the set $C$ : $C_1$**

*Definition 2:*The equation $c + c' - n = a + \beta_0$ has a solution in the numbers $c$, $c'$, $a$ where $c\notin C$, $c'\notin C$; $a\in A$; $c$, $c'$, $a \in (0,n)$.

Let $C^{*}$ be the set of all numbers $c,c'$ included in the above solution so that $C\cap C^{*}$ is the empty set.

We will call $C_1 = C \cup C^{*}$ the "canonical extension" of the set $C$.

**Canonical Extension of the set $B$ : $B_1$**

*Definition 3:*Let $B^{*}$ be the set of all values $\beta_0 + n - c$ where $c\in C^{*}$

Each number $b*\in B^{*}$ can be written in the form $c' - a$ where $c' \in C^{*}$ and $a \in A$ [Since $\beta_0+n- c = c' - a$ from the equation $c + c' - n = a + \beta_0$ above]

Since $b*$ has the form $\beta_0+n-c$, it follows that $b* \ge \beta_0 \ge 0$

Since $b* = c' - a$, $b* \le c' \le n$

So, all numbers $b*\in B^{*}$ lie in the interval $(0,n)$.

For all numbers $b*\in B^{*}$, $b*\notin B$ since otherwise, $b* = c' - a$ would imply that since $c' = b* + a$, $c' \in C$ which is false.

We will call $B_1 = B\cup B^{*}$ the "canonical extension" of the set $B$.

**: $A + B_1 = C_1$**

*Lemma 1*Proof:

(1) Either $b_1\in B$ or $b_1\in B^{*}$

(2) If $b_1 \in B$, then $a + b_1 \in C \subseteq C_1$

(3) If $b_1 \in B^{*}$, then $b_1 = \beta_0 + n - c'$ where $c' \notin C$

(4) Since $c + c' - n = a + \beta_0$, $b_1 = c - a$

QED