[see also: like, since
If f is as in (8), then...... [Not: “like in (8)'']
......, where each function g is as specified <described> 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.
As a first step we identify the image of Δ.
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≡ B as right modules.
Then E is irreducible as an L-module.
......, as is easily verified.
......, as noted <as was noted> in Section 2. [Not: “as it was noted”]
......, as desired <claimed/required>.
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). As for (4), this 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]