Yes. Define the combining function as follows:
takeP :: (a -> Bool) -> [a] -> [a]
takeP p = foldr step []
  where
    step x rest = if p x then x : rest else []
For the fusion derivation use the identity reconstruction foldr (:) [], take , and choose the displayed step as . The function is strict in the input list, , and its cons equation is exactly . Fusion gives the implementation. On a false test, step does not demand the folded tail, so the implementation can terminate after a finite list prefix even when the remaining list is infinite or undefined. An undefined predicate result at a tested element makes both definitions undefined.

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