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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