Relative derived subdivision makes the fixed subcomplex full

A simplex of $D_{A}K$ consists of a face $\alpha$ of $A$ and barycenters $b_{{\sigma }_1},\dots ,b_{{\sigma }_s}$ of nested faces outside $A$, strictly containing $\alpha$. The only vertices in $\mathrm{\mid }A\mathrm{\mid }$ are those of $\alpha$: a barycenter of $\sigma \mathrm{\notin }A$ has support $\sigma$, so cannot lie in the subcomplex $A$. Thus the vertices in $A$ span precisely $\alpha$.

For a point in that simplex with a positive coefficient at some $b_{{\sigma }_j}$, take the largest such face ${\sigma }_j$. All its vertex coordinates in the original simplex become positive, with no cancellation, so the original support contains ${\sigma }_j$. Such a point cannot belong to $\mathrm{\mid }A\mathrm{\mid }$, since that would put its support and every subface, including ${\sigma }_j$, in $A$. Conversely every point using only vertices of $\alpha$ lies in $\mathrm{\mid }A\mathrm{\mid }$. Therefore the intersection is exactly $\mathrm{\mid }\alpha \mathrm{\mid }$. The same argument applies with $K$ replaced by each $D_A^{r-1}K$. If $A$ is empty intersections are empty; if $A=K$, every simplex is already in $A$.