[see also: individual, particular, specific, unique, only]
This will be proved by showing that $H$ has but a single orbit on $M$.
Each row of $A$ has a single $\pm 1$ and the rest of the entries 0.
The two examples, $E_1$ and $E_2$, differ by only a single sequence, $e$, and they serve to illustrate the delicate nature of Theorem 2.
Can $E$ consist of a single point?
With this definition of a tree, no vertex is singled out as the root.
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