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 .
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 $M$.