Both gallery divisibility orders are lattices

The intersection $D(a)\cap D(b)$ contains the identity, is prefix closed and satisfies rank-two closure, since both principal ideals do. Length is bounded by $\min (\mathrm{\ell }(a),\mathrm{\ell }(b))$. The criterion gives a unique generator $d$, which is precisely the greatest common left divisor. Left divisibility is a partial order because additive length excludes nonidentity invertibles.

Choose one common right multiple $z$ of $a,b$, and let $U$ be the nonempty set of all common right multiples. The intersection $J={\bigcap }_{v\in U}D(v)$ is length bounded by $\mathrm{\ell }(z)$, contains the identity and satisfies prefix and rank-two closure. The criterion gives $J=D(m)$. Both $a,b$ belong to every $D(v)$, hence to $D(m)$, so $m\in U$. By definition $m$ divides every element of $U$, proving that $m$ is the least common right multiple. This argument does not assume that $U$ is finite.

Reversal sends incoming arrows ending at $A$ bijectively to outgoing arrows starting at $A$. If $b=ca$, then $b^{\ast }=a^{\ast }{c}^{\ast }$, and the converse follows by reversing again. Thus reversal preserves and reflects the specified divisibility relations. It is an order isomorphism, so it transports the meet and join just constructed to the incoming right-divisibility order. Thus both required orders are lattices.