问题
Is the following a definition of structural induction?
foldr f a (xs::ys) = foldr f (foldr f a ys) xs
Can someone give me an example of structural induction in Haskell?
回答1:
You did not specify it, but I will assume :: means list concatention and
use ++, since that is the operator used in Haskell.
To prove this, we will perform induction on xs. First, we show that the
statement holds for the base case (i.e. xs = [])
foldr f a (xs ++ ys)
{- By definition of xs -}
= foldr f a ([] ++ ys)
{- By definition of ++ -}
= foldr f a ys
and
foldr f (foldr f a ys) xs
{- By definition of xs -}
= foldr f (foldr f a ys) []
{- By definition of foldr -}
= foldr f a ys
Now, we assume that the induction hypothesis foldr f a (xs ++ ys) = foldr
f (foldr f a ys) xs holds for xs and show that it will hold for the list
x:xs as well.
foldr f a (x:xs ++ ys)
{- By definition of ++ -}
= foldr f a (x:(xs ++ ys))
{- By definition of foldr -}
= x `f` foldr f a (xs ++ ys)
^------------------ call this k1
= x `f` k1
and
foldr f (foldr f a ys) (x:xs)
{- By definition of foldr -}
= x `f` foldr f (foldr f a ys) xs
^----------------------- call this k2
= x `f` k2
Now, by our induction hypothesis, we know that k1 and k2 are equal,
therefore
x `f` k1 = x `f` k2
Thus proving our hypothesis.
来源：https://stackoverflow.com/questions/14960148/structural-induction-in-haskell