Empty Group Word is Reduced
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Although this article appears correct, it's inelegant. There has to be a better way of doing it.
Rather than mentioning them in passing and expecting the reader to understand what is meant, we need $w_i$ etc. to be defined. I know from experience that you can't rely on people involved in this website to have adequate knowledge, understanding or deductive abilities to be able to work it out for themselves.
You can help Proof Wiki by redesigning it.
Theorem
Let $S$ be a set
Let $\epsilon$ be the empty group word on $S$.
Then $\epsilon$ is reduced.
Proof
Rather than mentioning them in passing and expecting the reader to understand what is meant, we need $w_i$ etc. to be defined. I know from experience that you can't rely on people involved in this website to have adequate knowledge, understanding or deductive abilities to be able to work it out for themselves.
You can help Proof Wiki by redesigning it.
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By definition, a group word $w = w_1 \cdots w_i \cdots w_n$ is reduced if and only if $w_i \ne w_{i + 1}^{-1}$ for all $i \in \set {1, \ldots, n - 1}$, which is vacuously true for $\epsilon$.
$\blacksquare$