有两个自然数不相等的证据吗？

Question

Using Shapeless's Nat type, how can I derive evidence that two natural numbers are not equal?

This is what I have so far, but it only proves that a given Nat is not equal to 0. How can I prove that any two Nat values are not equal?

trait NEq[A <: Nat, B <: Nat] extends Serializable

object NEq {
  def apply[A <: Nat, B <: Nat](implicit neq: A != B): NEq[A, B] = neq

  type !=[A <: Nat, B <: Nat] = NEq[A, B]

  implicit def neq1[B <: Nat] = new !=[Succ[B], _0] {}
  implicit def neq2[B <: Nat] = new !=[_0, Succ[B]] {}
}

Answer 1

Using Shapeless's =:!= is the easiest way to accomplish this, but it's not very generalizable (what if you need to show that two numbers differ by one, etc.?), and often when I start out using it I end up switching to a custom type class.

Your NEq is only missing one piece—you've got the base cases, but you need the inductive step. You know that a Succ is never _0 , but you also know that two Succ s aren't the same if their contents aren't the same:

trait NEq[A <: Nat, B <: Nat] extends Serializable

object NEq {
  def apply[A <: Nat, B <: Nat](implicit neq: A != B): NEq[A, B] = neq

  type !=[A <: Nat, B <: Nat] = NEq[A, B]

  implicit def neq1[B <: Nat]: Succ[B] != _0 = new !=[Succ[B], _0] {}
  implicit def neq2[B <: Nat]: _0 != Succ[B] = new !=[_0, Succ[B]] {}

  implicit def neq3[A <: Nat, B <: Nat](implicit neq: A != B): Succ[A] != Succ[B] =
    new !=[Succ[A], Succ[B]] {}
}

This will work as expected.

Answer 2

Why not:

 def neq[A <: Nat, B <: Nat](implicit ev: A <:!< B) = ev

Example:

scala> neq[_1, _2]
res8: shapeless.<:!<[shapeless.nat._1,shapeless.nat._2] = shapeless.package$$anon$2@7db44dae

scala> neq[_1, _1]
<console>:15: error: ambiguous implicit values:
 both method nsubAmbig1 in package shapeless of type [A, B >: A]=> shapeless.<:!<[A,B]
 and method nsubAmbig2 in package shapeless of type [A, B >: A]=> shapeless.<:!<[A,B]
 match expected type shapeless.<:!<[shapeless.nat._1,shapeless.nat._1]
              neq[_1, _1]
                 ^

See also: Type constraint for type inequality in scala

And this , just in case if you want to check it in runtime for some reason.