as

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 [2], 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]

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