[*see also*: follow, entail, implicit]

Then $x=y$ and $y=z$ implies $x=z$. [*Or*: imply]

Note that $M$ being cyclic implies $F$ is cyclic.

However, if $B$ were omitted in (1), the case $n=0$ would imply $Nf=1$, an undesirable restriction.

Our present assumption implies that the last inequality in (8) must actually be an equality.

That (2) implies (1) is contained in the proof of Theorem 1 of [4].

The continuity of $f$ implies that of $g$.

We maintain the convention that implied constants depend only on $n$.

The equivalence of (a) and (b) is trivially implied by the definition of $M$.

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