[see also: contrary]

We must have $Lf=0$, for otherwise we can replace $f$ by $f-Lf$.

We claim that $f(z)>1$. Otherwise, the disc $D$ would intersect $B$.

We now prove ...... Indeed, suppose otherwise. Then ......

Unless otherwise stated, we assume that ......

Moreover, for $L$ tame or otherwise, it may happen that $E$ is a free module.

Simplicity (or otherwise) of the underlying graphs will be discussed in the next section.

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