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Then using the inductive hypothesis, (8), and (7), we have b ∈ A ( m − 1 ).
If there is some α i = 0, then delete the number α i and for the remaining n − 1 number, inequality (3.1) is obvious by using the inductive hypothesis.
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In order to finish this proof, we need to transform the goal to use the inductive hypothesis.
Now, we assume that holds for some fixed p. Then, using and the inductive hypothesis, we get d ( x n - 1, x n + p ) ≤ d ( x n - 1, x n ) + d ( x n, x n + p ) < δ + ε. (6).
However, data driven exploration of the space is possible using the Inductive Causation (IC) algorithm [ 11].
Responses were analysed using the inductive thematic analysis procedure described by Hayes [ 19].
This logic can then be extended for β ≥ 1 using the inductive arguments described previously.
We can then utilize the inductive hypothesis to complete the proof.
Now that we are remained with ℓ erasures, we can employ the inductive hypothesis to complete the proof.
From formula (A3) and the inductive hypothesis, and using n≥2, we can write (A5) Now using formulae (A2) and (A5), we get and from the inductive hypothesis and using n≥2, we can verify that ɛ i +1 satisfies Since we have, then, by induction, we have, for any integer k, and thus, with n≥2, Putting it all together, we have Moreover, from formulae (A1) and (A3), we can write ▪.
It is easy to see (using (2.8), (2.9) and the inductive hypothesis) that the last maximum is equal to D ( z, T z ), i.e., D ( T n + 1 z, T z ) ≤ r K D ( z, T z ) and relation (2.10) is proved by induction.
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.

Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com