Question

假设您有以下定义：

abstract class IntSet {
  def incl(x: Int): IntSet    
  def contains(x: Int): Boolean    
  def union(other: IntSet): IntSet
}

case class NonEmpty(elem: Int, left: IntSet, right: IntSet) extends IntSet {
  def incl(x: Int) =
    if (x < elem) NonEmpty(elem, left incl x, right)
    else if (x > elem) NonEmpty(elem, left, right incl x)
    else this

  def contains(x: Int) =
    if (x < elem) left contains x
    else if (x > elem) right contains x
    else true

  def union(other: IntSet) = (left union (right union other)) incl elem
}

object Empty extends IntSet {
  def incl(x: Int) = NonEmpty(x, Empty, Empty)    
  def contains(x: Int) = false    
  def union(other: IntSet) = other
}

并且必须证明以下命题：

(xs union ys) contains x = xs contains x || ys contains x

从这里我推断出两个基本案例。 xs =空，ys =空。这是我因为以下原因而陷入困境的第二个基本案例：

// substituting ys = Empty
 (xs union Empty) contains x = xs contains x || Empty contains x
// RHS:
 xs contains x || false
 xs contains x
// LHS:
 ((left union (right union Empty)) incl elem) contains x // By definition of NonEmpty.union

如何将LHS减少到xs包含x？我是否必须对xs union Empty = xs做另一个归纳假设，如果是，那怎么能用于表达式呢？

Answer 1

证明：

(xs union ys) contains x = xs contains x || ys contains x

给出：

以上定义
Empty contains x = false
(s incl x) contains x = true
(s incl y) contains x = s contains x ; if x != y

仅使用xs而不是ys的归纳步骤和结构证明，请参考原始问题

案例1：

if xs = Empty

  左手边：
  (Empty union ys) contains x
  = ys contains x => Definition of Empty union other

右手侧：
  Empty contains x || ys contains x
  = false || ys contains x => Definition of Empty contains x
  = ys contains x => Truth table of OR ( false OR true = true, false OR false = false)

  左手边=右手边，因此证明了归纳法，该语句对所有子树都是正确的

第二种情况：
  if (xs is NonEmpty(z, l, r) and z = x)

  左手边：
  (NonEmpty(x, l, r) union ys) contains x
  = ((l union(r union ys)) incl x) contains x => From definition of union on NonEmpty
  = true => from (3) above (s incl x) contains x = true

  右侧：
  xs contains x || ys contains x
  = NonEmpty(x, l, r) contains x || ys contains x => From definition of xs
  = true || ys contains x => From definition of contains on NonEmpty
  = true => Truth table of OR (true OR false = true, true OR true = true)

左侧=右侧

案例III：
  if ( xs is NonEmpty(z, l, r) and z < x )

  左手边：
  (NonEmpty(z, l, r) union ys) contains x
  = (l union(r union ys) incl z) contains x => From definition of union
  = (l union(r union ys)) contains x => From (4) above (s incl x) contains y = s contains y; if x != y
  = (l contains x || (r union ys) contains x) => From induction step, the statement is true for all subtrees
  = (l contains x || r contains x || ys contains x) => From induction step, the statement is true for all subtrees
  = r contains x || ys contains x => since z < x, definition of contains (l contains x) returns false

  右侧：
  (if z < x)
  NonEmpty(z, l, r) contains x || ys contains x
  = r contains x || ys contains x => definition of contains,

  左侧=右侧

案例if ( xs is NonEmpty(z, l, r) and z > x )与上面的案例III类似由此证明

IntSet联合证明结构归纳

1 个答案: