topological sort equivalence
Let $\tau_1: x_1, x_2, \dots, x_n$ and $\tau_2: y_1, y_2, \dots, y_n$ be two topological orderings of the vertices of graph $G$. Our aim is to show that $\tau_1$ can be transformed into $\tau_2$ by a sequence of swaps, each changing the position of two vertices in the sequence, not connected to each other by an edge of $G$.

Starting from the beginning compare the vertices $x_i$ and $y_i$ and select the first pair that differs. For notational convenience we assume $x_1 \neq y_1$. As $y_1$ must occurs in $\tau_1$ we can find $x_j=y_1$, $j&gt;1$. Thus vertices $x_1, \dots, x_{j-1}$ are all before $x_j=y_1$ in ordering $\tau_1$, but all after $y_1=x_j$ in ordering $\tau_2$. This implies there is no edge in $G$ (in either direction) from $x_j$ to each of $x_1, \dots, x_{j-1}$. Thus we can "legally" swap $x_j$ with each of $x_1, \dots, x_{j-1}$ to get it to the first position. We obtain $\tau'_1: x_j, x_1, \dots, x_{j-1}, x_{j+1}, \dots, x_n$ which has the same first element as $\tau_2$.

Iterate, or invoke induction.

(Thanks for your suggestions and support.)