プログラミングhaskell 13章 問題7
DESCRIPTION
スタートHaskellでの発表TRANSCRIPT
(定義1) map f [] = [](定義2) map f (x:xs) = f x : map f xs(定義3) (f . g) x = f (g x)
map f (map g xs) = map (f . g) xs
map f (map g []) = map (f . g) []
map f (map g []) = map (f . g) []map f [] = []
(定義1) map f [] = []
map f (map g []) = map (f . g) []map f [] = [][] = []
(定義1) map f [] = []
(仮定1) map f (map g xs) = map (f . g) xs
map f (map g (x:xs)) = map (f . g) (x:xs)
map f (map g (x:xs)) = map (f . g) (x:xs)map f (g x : map g xs) = (f . g) x : map (f . g) xs
(定義2) map f (x:xs) = f x : map f xs
map f (map g (x:xs)) = map (f . g) (x:xs)map f (g x : map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = (f . g) x : map (f . g) xs
(定義2) map f (x:xs) = f x : map f xs
map f (map g (x:xs)) = map (f . g) (x:xs)map f (g x : map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = f (g x) : map (f . g) xs
(定義3) (f . g) x = f (g x)
map f (map g (x:xs)) = map (f . g) (x:xs)map f (g x : map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = f (g x) : map (f . g) xsf (g x) : map (f . g) xs = f (g x) : map (f . g) xs
(仮定1) map f (map g xs) = map (f . g) xs
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