Since the join of simplicial sets is associative and $\Delta^m = \Delta^0 \star \cdots \star \Delta^0$ ($m+1$ times), we should start by trying to understand things like $\Lambda^n_j \star \Delta^0$, a.k.a. the "final" cone on $\Lambda^n_j$. It's not too hard to see that this is the subcomplex of $\Delta^{n+1}$ consisting of those faces which do not contain the (codimension 2) face $\{0, \ldots, r-1, r+1, \ldots, n\}$. In other words, we are missing the face opposite $r$ and $n+1$, because we were originally missing the face opposite $r$ of $\Delta^n$, as well as the three other faces (including the interior of $\Delta^{n+1}$) it contains. Similarly $\Delta^0 \star \partial \Delta^n$ is the horn $\Lambda^{n+1}_0$ (we are missing the interior of $\Delta^{1,\ldots,n}$ and the cone on it).

In general all the simplicial sets that come up have the form of the subcomplex of $\Delta^N$ consisting of those faces which do not contain a fixed face $\Delta^S$, $S \subset \{0, \ldots, n\}$. Forming the cone (on either side) on such a space results in another such space with $N$ replaced by $N+1$ and $S$ unchanged (as a subset of the vertices of the original $\Delta^N$, which if we formed a cone on the left, means we increment each index in $S$).

After doing these sorts of computations, I expect that $\Lambda^n_j \star \Delta^m$ and $\Delta^n \star \partial \Delta^m$ will be two subcomplexes of $\Delta^{n+m+1}$ each characterized by avoiding faces containing a certain face, and that $\Lambda^n_j \star \partial \Delta^m$ is their intersection and $\Lambda^{n+m+1}_j$ is their union, from which the claim would follow.

Since the join of simplicial sets is associative and $\Delta^m = \Delta^0 \star \cdots \star \Delta^0$ ($m+1$ times), we should start by trying to understand things like $\Lambda^n_j \star \Delta^0$, a.k.a. the "final" cone on $\Lambda^n_j$. It's not too hard to see that this is the subcomplex of $\Delta^{n+1}$ consisting of those faces which do not contain the (codimension 2) face $\{0, \ldots, r-1, r+1, \ldots, n\}$. In other words, we are missing the face opposite $r$ and $n+1$, because we were originally missing the face opposite $r$ of $\Delta^n$, as well as the three other faces (including the interior of $\Delta^{n+1}$) it contains. Similarly $\Delta^0 \star \partial \Delta^n$ is the horn $\Lambda^{n+1}_0$ (we are missing the interior of $\Delta^{1,\ldots,n}$ and the cone on it).

In general all the simplicial sets that come up have the form of the subcomplex of $\Delta^N$ consisting of those faces which do not contain a fixed face $\Delta^S$, $S \subset \{0, \ldots, N\}$. Forming the cone (on either side) on such a space results in another such space with $N$ replaced by $N+1$ and $S$ unchanged (as a subset of the vertices of the original $\Delta^N$, which if we formed a cone on the left, means we increment each index in $S$).

After doing these sorts of computations, I expect that $\Lambda^n_j \star \Delta^m$ and $\Delta^n \star \partial \Delta^m$ will be two subcomplexes of $\Delta^{n+m+1}$ each characterized by avoiding faces containing a certain face, and that $\Lambda^n_j \star \partial \Delta^m$ is their intersection and $\Lambda^{n+m+1}_j$ is their union, from which the claim would follow.

Since the join of simplicial sets is associative and $\Delta^m = \Delta^0 \star \cdots \star \Delta^0$ ($m+1$ times), we should start by trying to understand things like $\Lambda^n_j \star \Delta^0$, a.k.a. the "final" cone on $\Lambda^n_j$. It's not too hard to see that this is the subcomplex of $\Delta^{n+1}$ consisting of those faces which do not contain the (codimension 2) face $\{0, \ldots, r-1, r+1, \ldots, n\}$. In other words, we are missing the face opposite $r$ and $n+1$, because we were originally missing the face opposite $r$ of $\Delta^n$, as well as the three other faces (including the interior of $\Delta^{n+1}$) it contains. Similarly $\Delta^0 \star \partial \Delta^n$ is the horn $\Lambda^{n+1}_0$ (we are missing the interior of $\Delta^{1,\ldots,n}$ and the cone on it).

After doing these sorts of computations, I expect that $\Lambda^n_j \star \Delta^m$ and $\Delta^n \star \partial \Lambda^m$ will be two subcomplexes of $\Delta^{n+m+1}$ each characterized by avoiding faces containing a certain face, and that $\Lambda^n_j \star \partial \Delta^m$ is their intersection and $\Lambda^{n+m+1}_j$ is their union, from which the claim would follow.

Since the join of simplicial sets is associative and $\Delta^m = \Delta^0 \star \cdots \star \Delta^0$ ($m+1$ times), we should start by trying to understand things like $\Lambda^n_j \star \Delta^0$, a.k.a. the "final" cone on $\Lambda^n_j$. It's not too hard to see that this is the subcomplex of $\Delta^{n+1}$ consisting of those faces which do not contain the (codimension 2) face $\{0, \ldots, r-1, r+1, \ldots, n\}$. In other words, we are missing the face opposite $r$ and $n+1$, because we were originally missing the face opposite $r$ of $\Delta^n$, as well as the three other faces (including the interior of $\Delta^{n+1}$) it contains. Similarly $\Delta^0 \star \partial \Delta^n$ is the horn $\Lambda^{n+1}_0$ (we are missing the interior of $\Delta^{1,\ldots,n}$ and the cone on it).

In general all the simplicial sets that come up have the form of the subcomplex of $\Delta^N$ consisting of those faces which do not contain a fixed face $\Delta^S$, $S \subset \{0, \ldots, n\}$. Forming the cone (on either side) on such a space results in another such space with $N$ replaced by $N+1$ and $S$ unchanged (as a subset of the vertices of the original $\Delta^N$, which if we formed a cone on the left, means we increment each index in $S$).

After doing these sorts of computations, I expect that $\Lambda^n_j \star \Delta^m$ and $\Delta^n \star \partial \Delta^m$ will be two subcomplexes of $\Delta^{n+m+1}$ each characterized by avoiding faces containing a certain face, and that $\Lambda^n_j \star \partial \Delta^m$ is their intersection and $\Lambda^{n+m+1}_j$ is their union, from which the claim would follow.