[see also: like, since]
If $f$ is as in (8), then ...... [Not: “like in (8)”]
We can multiply two elements of $E$ by concatenating paths, much as in the definition of the fundamental group.
......, where each function $g$ is as specified $\langle$described$\rangle$ above.
Actually, [3, Theorem 2] does not apply exactly as stated, but its proof does.
They were defined directly by Lax , essentially as we have defined them.
For $k=2$ the count remains as is.
In the case where $A$ is commutative, as it will be in most of this paper, we have ......
As a first step we identify the image of $\Delta$.
Then $F$ has $T$ as its natural boundary.
The algorithm returns 0 as its answer.
Now $X$ can be taken as coordinate variable on $M$.
If one thinks of $x, y$ as space variables and of $z$ as time, then ......
Then $G$ is a group with composition as group operation.
We have $A\equiv B$ as right modules.
Then $E$ is irreducible as an $L$-module.
......, as is easily verified.
......, as noted $\langle$as was noted$\rangle$ in Section 2. [Not: “as it was noted”]
......, as desired $\langle$claimed/required$\rangle$.
The elements of $F$ are not in $S$, as they are in the proof of ......
Note that $F$ is only nonnegative rather than strictly positive, as one may have expected.
Then $G$ has 10 normal subgroups and as many non-normal ones.
Moreover, $H$ is a free $R$-module on as many generators as there are path components of $X$.
But $A$ has three times as many elements as $B$ has.
We can assume that $p$ is as close to $q$ as is necessary for the following proof to work.
Then $F$ can be as great as 16.
Each tree is about two-thirds as deep as it was before.
As $M$ is ordered, we have no difficulty in assigning a meaning to $(a,b)$.
The ordered pair $(a,b)$ can be chosen in 16 ways so as not to be a multiple of $(c,d)$.
Now (3) is clear. As for (4), it is an immediate consequence of Lemma 6. [= Concerning (4)]
As with the digit sums, we can use alternating digit sums to prove ...... [= Just as in the case of digit sums]
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