Kruskal-Katona theorem: Why shifting preserves the neighborhood?

Recall that a neighbor of a binary vector $v$ is a vector which can be obtained from $v$ by flipping one of its $1$-entries to $0$. A shadow of a set $B\subseteq\{0,1\}^n$ of vectors is the set $\partial(B)$ of all its neighbors. The $j$-th shift of a vector $v\in B$ is defined by: \[ s_j(v)=\begin{cases} v\oplus e_1\oplus e_j & \mbox{if $v_1=0$, $v_j=1$ and $v\oplus e_1\oplus e_j\not\in B$}\\ v & \mbox{otherwise} \end{cases} \] Note that, due to the condition $v\oplus e_1\oplus e_j\not\in B$, the definition of shifts depends on what set $B$ of vectors we are dealing with. Note also that \begin{equation} \mbox{$s_j(v)=v$ if either $v_1=1$ or $v_1=v_j=0$.}\tag{i} \end{equation} The $j$-th shift of $B$ is $s_j(B)=\{s_j(v)\colon v\in B\}$.

Claim: Neighbors of shits are shifts of neighbors. That is, for every every $1 < j\leq n$, \[ \partial(s_j(B))\subseteq s_j(\partial(B))\,. \]

Proof: We have to show that for every $v\in B$, $\partial(s_j(v))\subseteq s_j(\partial(B))$ holds. So let $u$ be a neighbor of the $j$-th shift $s_j(v)$ of $v$. It is enough to show that \begin{equation} \mbox{$u=s_j(w)$ for some neighbor $w$ of $v$.} \tag{ii} \end{equation} In all the case below, except for the last one, (ii) will hold for $w=u$.

We first consider the case when $s_j(v)=v$, that is, when either $v_1=v_j=0$ or $v_1=1$. Then $u=v\oplus e_k$ is a neighbor of $v$ for some $k\in\{1,\ldots,n\}$. If $k\neq 1$, then (ii) holds with $w=u$. If $k=1$, then $v_1=1$.

Case $k=1$ and $v_j=0$

In this case, (ii) again holds with $w=u$. \[ \begin{array}{rccccc} & & 1 & & j & \\ v =s_j(v) & = & 1 & \cdots\cdots & 0 & \cdots\cdots \\ u & = & 0 & \cdots\cdots & 0 & \cdots\cdots \\ \end{array} \]

Case $k=1$ and $v_j=1$

In this case, vector $u\oplus e_1\oplus e_j$ is a neighbor of $v$, and hence, belongs to $\partial(B)$. By the definition of the shift, this implies that $s_j(u)=u$. Since, $u$ is a neighbor of $s_j(v)=v$, (ii) again holds with $w=u$. \[ \begin{array}{rccccc} & & 1 & & j & \\ v =s_j(v) & = & 1 & \cdots\cdots & 1 & \cdots\cdots \\ u & = & 0 & \cdots\cdots & 1 & \cdots\cdots \\ u\oplus e_1\oplus e_j& = & 1 & \cdots\cdots & 0 & \cdots\cdots \\ \end{array} \]

Suppose now that $s_j(v)\neq v$, that is, $v_1=0$, $v_j=1$ and $s_j(v)=v\oplus e_1\oplus e_j$. Let $u< s_j(v)$ be a neighbor of $s_j(v)$.

Case $u_1=0$

In this case $u$ is also a neighbor of the vector $v\in B$ and, by (i), is a shift of this neighbor (of itself), because $s_j(u)=u$. That is, in this case (ii) holds with $w=u$. \[ \begin{array}{rccccc} & & 1 & & j & \\ v& = & 0 & \cdots\cdots & 1 & \cdots\cdots \\ s_j(v)& = & 1 & \cdots\cdots & 0 & \cdots\cdots \\ u & = & 0 & \cdots\cdots & 0 & \cdots\cdots \\ \end{array} \]

Case $u_1=1$

In this case, $u$ is obtained form $s_j(v)$ by flipping some $k$-th bit $k\not\in\{1,j\}$ from $1$ to $0$. Since $u_1=1$, we have that $s_j(u)=u$. Hence, if $u\in\partial(B)$, then (ii) holds with $w=u$.
So assume that $u\not\in\partial(B)$. The vector $x=v\oplus e_k$ is a neighbor of $v$. Moreover, $x\oplus e_1\oplus e_j=u\not\in\partial(B)$. Since $x_1=0$ and $x_j=1$, we then have that $s_j(x)=x\oplus e_1\oplus e_j=u$ (by the definition of the shift). Hence, in this case, (ii) holds with $w=x$. \[ \begin{array}{rccccccc} & & 1 & & j & & k & \\ v & = & 0 & \cdots\cdots & 1 & \cdots\cdots & 1 & \cdots\cdots \\ s_j(v) & = & 1 & \cdots\cdots & 0 & \cdots\cdots & 1 & \cdots\cdots \\ u & = & 1 & \cdots\cdots & 0 & \cdots\cdots & 0 & \cdots\cdots \\ x & = & 0 & \cdots\cdots & 1 & \cdots\cdots & 0 & \cdots\cdots \\ x\oplus e_1\oplus e_j & = & 1 & \cdots\cdots & 0 & \cdots\cdots & 0 & \cdots\cdots \\ \end{array} \]