Why can't a method be defined both for a struct and its pointer?

Given the setup in the 54th slide of the golang tour:

type Abser interface {
	Abs() float64
}
type Vertex struct {
	X, Y float64
}
func (v *Vertex) Abs() float64 {
	return math.Sqrt(v.X*v.X + v.Y*v.Y)
}

Why can't a method also be defined for the struct as well as the pointer to the struct? That is:

func (v Vertex) Abs() float64 {
	return math.Sqrt(v.X*v.X + v.Y*v.Y)
}

Defining this gives the following error:

prog.go:41: method redeclared: Vertex.Abs
	method(*Vertex) func() float64
	method(Vertex) func() float64

It can. Just define it on the struct and not the pointer. It will resolve both ways

Method Sets
> The method set of the corresponding pointer type *T is the set of all
> methods with receiver *T or T (that is, it also contains the method
> set of T)

Try live: http://play.golang.org/p/PsNUerVyqp

package main
import (
	"fmt"
	"math"
	)
type Abser interface {
	Abs() float64
}
type Vertex struct {
	X, Y float64
}
func (v Vertex) Abs() float64 {
	return math.Sqrt(v.X*v.X + v.Y*v.Y)
}
func main() {
	v := Vertex{5, 10}
	v_ptr := &v
	fmt.Println(v.Abs())
	fmt.Println(v_ptr.Abs())
}

Update: As per comments I have created an extra example that actually makes use of the Abser interface to illustrate that both the value and the pointer satisfy the interface.

https://play.golang.org/p/Mls0d7_l4_t

While considering for example:

type T U
func (t *T) M() int { return 1 }
var t T

...we can now invoke M() on t by writing t.M() as the language permits to call a method with a pointer receiver even on its underlying (non pointer) typed instances, i.e. it becomes equivalent to (&t).M().

If it will be permitted to now additionaly define:

func (t T) M() int { return 2 }

...then there's no way to tell what is now t.M() supposed to return.