unf和list的连接是unf - 证明Haskell

Question

如何证明每个列表xs都符合以下条件：

undefined ++ xs = undefined

Answer 1

没有太多要证明的。只有一条规则（无法解释或分解为更小的规则）case语句尝试将undefined与构造函数匹配，导致undefined。一旦接受此规则，我们就可以观察

undefined ++ ys
= { by definition of ++ }
case undefined of
    [] -> ys
    x:xs -> x : (xs ++ ys)
= { case that matches undefined against a constructor }
undefined