The document proves that map f (map g xs) = map (f . g) xs for any functions f and g and list xs. It does this by: 1) Starting from the base case that mapping an empty list results in an empty list 2) Assuming map f (map g xs) = map (f . g) xs is true for some list xs 3) Showing through substitutions and definitions that this holds true for lists of the form x:xs as well 4) Concluding that the original statement is true for all lists xs

2. (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

3. map f (map g []) = map (f . g) []

4. map f (map g []) = map (f . g) []
map f [] = []
(定義1) map f [] = []

5. map f (map g []) = map (f . g) []
map f [] = []
[] = []
(定義1) map f [] = []

6. (仮定1) map f (map g xs) = map (f . g) xs
map f (map g (x:xs)) = map (f . g) (x:xs)

7. 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

8. 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
f (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

9. 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
f (g x) : map f (map g xs) = (f . g) x : map (f . g) xs
f (g x) : map f (map g xs) = f (g x) : map (f . g) xs
(定義3) (f . g) x = f (g x)

10. 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
f (g x) : map f (map g xs) = (f . g) x : map (f . g) xs
f (g x) : map f (map g xs) = f (g x) : map (f . g) xs
f (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

11. http://www.paraiso-lang.org/ikmsm/