Haskell - prove by induction

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予麋鹿
予麋鹿 2021-02-14 23:09

where im given these definitions

(++)::[a]->[a]
[]++ys = ys
(x:xs)++ys = x:(xs++ys)

filter::(a->Bool)->[a]->[a]:
filter p [] = []
filter p (x:xs) = i         


        
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