Question

我获得了以下2个版本的树平展程序，我被要求证明相同的行为。

flatten::Tree->[Int]
flatten (Leaf z)        =[z]                             F.1
flatten (Node x y)      =concat (flatten x)(flatten y)   F.2

flatten'::Tree->Int->[Int]
flatten' (Leaf z) a     =concat [z] a                    P.1
flatten' (Node x y) a   =flatten' x (flatten' y a)       P.2

concat->[a]->a->[a]
concat [] a             =a                               C.1
concat (h:t) a          =h:concat t a                    C.2

证明：

flatten' z a = concat (flatten z) a

基本案例：

LHS:
    flatten' (Leaf z) a = concat [z] a                   By P.1
RHS:
    concat (flatten z) a = concat (flatten (Leaf z)) a
                         = concat [z] a                  By F.1 and C.1

LHS=RHS, hence base case holds

归纳案例：

（唯一可能的感谢下面的那个人解释了感应如何二叉树工作！）

Assume that:
   flatten' x a = concat (flatten x) a                   Ind. Hyp1
   flatten' y a = concat (flatten y) a                   Ind. Hyp2

Show that:
   flatten'(Node x y) a = concat (flatten (Node x y)) a

LHS:
   flatten' (Node x y) a = 
   flatten' x (flatten' y a)                             By P.2
   flatten' x (concat (flatten y) a)                     By Ind. Hyp2
   concat (flatten x) (concat (flatten y) a)             By Ind. Hyp1

RHS:
   concat (flatten (Node x y)) a =
   concat (concat (flatten x) (flatten y)) a             By F.2
   concat (flatten x) (concat (flatten y) a)             By C.2

   LHS = RHS, hence inductive step holds. End of proof.

Answer 1

当引入列表时，您的归纳假设是所需属性保留在列表尾部，并且您必须证明它也保留在整个列表中。

在树上，它只是略有不同：你的归纳假设是，想要的属性保持在两个子树上，你必须证明它也适用于整个树。

Assume that:
   forall a, flatten' x a = concat (flatten x) a         Ind. Hyp. 1
   forall a, flatten' y a = concat (flatten y) a         Ind. Hyp. 2

Show that:
   forall a, flatten'(Node x y) a = concat (flatten (Node x y)) a

我想你现在可以猜到如何从这里开始，所以我不会破坏这种乐趣。您可能需要依赖concat的某些基本属性来执行某些子步骤。

最后注意事项：在你的基本案例中，你提到C.1作为理由 - 你确定你真的使用过它吗？

结构感应证明的归纳步骤

1 个答案: