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Before continuing the proof, we give the following lemma that will be used later.
Then it follows from (3.11) that for any function and for all β ∈ [ 0, 27 4 ), ∥ P 0 u ∥ L 2 ( R + ; H ) 2 − β ∥ A 3 − j d j u d t j ∥ L 2 ( R + ; H ) 2 > 0. Passing here to the limit as β → 27 4, we obtain that n j ≤ 2 3 3, and hence, n j = 2 3 3. Continuing the proof of the theorem, we suppose that n j > 2 3 3. Then n j − 2 ∈ ( 0, 27 4 ).
Similar(58)
We continue the proof of Stillman's question due to Ananyan-Hochster.
Let us continue the proof.
We continue the proof of the theorem.
To continue the proof, we need the following notation.
end{aligned} We continue the proof with several cases.
Hence, one may continue the proof as in the first subcase.
When we continue the proof of the general case, we will work with the sequence the sequence instead of (f n ).
To continue the proof of Theorem 1, consider the function ξ : q ↦ a q taking each q ∈ Q to the first positive zero of the function y q.
However, Cohen continued, the proof was in deeds as well as words: "After seeing John in action for a year, the board decided to award him the additional post".
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com