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

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

(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 $a^{\mathrm{\prime }}\in A$ with $a^{\mathrm{\prime }}>a$, whence $a^{\mathrm{\prime }}+b\in A+B$ and $a^{\mathrm{\prime }}+b>a+b=s$.

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

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

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

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

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

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