Note that $M$ being cyclic implies $F$ is cyclic.
The probability of $X$ being rational equals 1/2.
This is exactly our definition of a weight being regulated.
We have to show that the property of there being $x$ and $y$ such that $x<y$ uniquely determines $P$ up to isomorphism.
In addition to $f$ being convex, we require that ......
Here $J$ is defined to equal $Af$, the function $f$ being as in (3). [= where the function $f$ is ......; not: “being the function $f$ as in”]
......, the constant $C$ being independent of ......
The ideal is defined by $m=$ ......, it being understood that ......
But ......, it being impossible to make $A$ and $B$ intersect. [= since it is impossible to make]
The map $F$ being continuous, we can assume that ......
Actually, $S$ has the much stronger property of being convex.
This method has the disadvantage of not being intrinsic.
[Do not write: “the function $f$ being the solution of (1)” if you mean: the function $f$ that is the solution of $(1)$.]
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