Bruns & Herzog (Cohen-Macaulay Rings) give the following definition of a pure shellable simplicial complex:
I am stuck in their proof that condition $(b) \implies (c)$:
In the argument depicted above $S_i=\left\{F: F \in \langle F_1,\dots,F_i \rangle, \, F \not\in \langle F_1,\dots,F_{i-1} \rangle \right\}.$
Question 1: Am I correct that $S_i$ is the same set as $\left\{F: F \subset F_i, F \not\subset F_j, \forall j<i\right\}$?
Question 2: I don't see how they arrive at the conclusion that $F_i \setminus F_k = \left\{v\right\}$. If i am reading this right, it says that every element of $F_i$ is an element of $F_k$ except from $v \in F_i$. What we know about $G$ is that it is the unique minimal face of $F_i$ that it not a face of any of the $F_1,\dots,F_{i-1}$. So if we remove $v$ from $G$, we get a face of say $F_k$. But how does that imply that $F_i \setminus F_k = \left\{v\right\}$?
Proof. If $w\in F_i-\{v\}$, and $w\notin F_k$, then $w\in F_i-F_k$, a contradiction. For conversely note that $v\notin F_k$, otherwise $F_i\subseteq F_k$ and therefore $F_i=F_k$ which is impossible. If $w\in F_i-F_k$ and $w\ne v$, then $w\in F_i-\{v\}$, so $w\in F_k$, a contradiction.
Now let $G$ be the unique minimal element of $S_i=\{F:F\subset F_i\text{ and } F\nsubseteq F_j\text{ for all }j<i\}$, and $v\in G-F_j$. If $F_i-F_k\ne \{v\}$ for all $k<i$, then $F_i-\{v\}\nsubseteq F_k$ for all $k<i$, so $F_i-\{v\}\in S_i$. Since every element of $S_i$ contains $G$, we must have $G\subseteq F_i-\{v\}$, a contradiction.