no

This set has no fewer elements than $K$ has.

Then $X$ has no fewer than twenty elements.

The case $n=2$ is of no interest since ......

For $F=R$, no integration over $M$ is needed in (5).

Thus $F$ has no pole in $U$ (hence none in $V$).

No $x$ has more than one inverse.

There is no map such that ......

For no $x$ does the limit exist. [Note the inversion after the negative clause.]

No two members of $A$ have an element in common.

We conclude that, no matter what the class of $b$ is, we have an upper bound on $M$.



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