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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