代数数据类型的计数

我正在阅读/收听 Chris Taylor 关于代数数据类型的演示。

http://chris-taylor.github.io/blog/2013/02/10/the-algebra-of-algebraic-data-types/

还有一个关于函数类型的部分。具体例子

data Bool = True | False
data Trio = First | Second | Third

鉴于法律规定

a -> b == B^A

Given

Trio -> Bool     should equal     8

为什么乘法是 8 而不是 6？

如果我理解正确的话，具体组合应该是

First  -> True
First  -> False
Second -> True
Second -> False
Third  -> True
Third  -> False

这不就是6个具体的实现吗Trio -> Bool?

我缺少什么？

Those aren't full implementations. For the full implementations, it is like counting from 0 to 7 (which is a total of 8 = 2³ numbers) in binary, with each line of each implementation representing one of the three bits. All the possibilities look like this (if we call our function f):

1)

f First  = False
f Second = False
f Third  = False

2)

f First  = True
f Second = False
f Third  = False

3)

f First  = False
f Second = True
f Third  = False

4)

f First  = True
f Second = True
f Third  = False

5)

f First  = False
f Second = False
f Third  = True

6)

f First  = True
f Second = False
f Third  = True

7)

f First  = False
f Second = True
f Third  = True

8)

f First  = True
f Second = True
f Third  = True