induced from those of $\mathcal{B}$.

\(ii) The composition in each level of $\mathcal{B}$ induces $$\mathcal{B}(x,y) \times \mathcal{B}(y,z)\to \mathcal{B}(x,z).$$ Similarly the identity map in $\mathcal{B}(x,x)$ is defined as $id_x$, the identity at $x$ in the category $\mathcal{B}_0$.

\(iii) The resulting structure is an $\mathcal{S}$-enriched category, that will also be denoted $\mathcal{B}$.$\blacksquare$

The proof is easy. In particular, this shows that $S(\mathbb{A})$ is a simplicially enriched category. The description of the simplices in each dimension of $S(\mathbb{A})$ that start at $a$ and end at $b$ is intuitively quite simple. The arrows in the category, $T(\mathbb{A})$ correspond to strings of symbols representing non-identity arrows in $\mathbb{A}$ itself, those strings being ‘composable’ in as much as the domain of the $i^{th}$ arrow must be the codomain of the $(i-1)^{th}$ one and so on. Because of this we have:\
$S(\mathbb{A})_0$ consists exactly of such composable chains of maps in $\mathbb{A}$, none of which is the identity;\
$S(\mathbb{A})_1$ consists of such composable chains of maps in $\mathbb{A}$, none of which is the identity, together with a choice of bracketting;\
$S(\mathbb{A})_2$ consists of such composable chains of maps in $\mathbb{A}$, none of which is the identity, together with a choice of two levels of bracketting;\
and so 