Cut addition: $A+B$ is a cut, commutative and associative, with identity $0^{*}$

(C1) $A + B$ is proper and nonempty: choosing $a \in A$, $b \in B$ gives $a + b \in A + B$, so $A + B \neq \varnothing$; choosing $a' \notin A$, $b' \notin B$, every $a \in A$, $b \in B$ satisfies $a < a'$ and $b < b'$, hence $a + b < a' + b'$, so $a' + b' \notin A + B$ and $A + B \neq \mathbb{Q}$.

(C2) $A + B$ is downward closed: if $s = a + b \in A + B$ with $a \in A$, $b \in B$, and $q < s$, then $q - a < b$, so $q - a \in B$ by (C2) for $B$; hence $q = a + (q - a) \in A + B$.

(C3) $A + B$ has no greatest element: given $s = a + b \in A + B$, (C3) for $A$ yields $a' \in A$ with $a' > a$, whence $a' + b \in A + B$ and $a' + b > a + b = s$.

Commutativity and associativity descend from $\mathbb{Q}$: $A + B = \{a + b\} = \{b + a\} = B + A$, and $(A + B) + C = \{(a + b) + c\} = \{a + (b + c)\} = A + (B + C)$.

$A + 0^{*} \subseteq A$: for $a \in A$ and $q \in 0^{*}$ (so $q < 0$), $a + q < a$, hence $a + q \in A$ by (C2).

$A \subseteq A + 0^{*}$: given $a \in A$, (C3) supplies $r \in A$ with $r > a$; then $a - r < 0$, so $a - r \in 0^{*}$, and $a = r + (a - r) \in A + 0^{*}$.

$A + B$ satisfies (C1)–(C3), so it is a Dedekind cut.

The two inclusions give the identity law $A + 0^{*} = A$.

Hence $A + B$ is a cut, and cut addition is commutative and associative with two-sided identity $0^{*}$.