Why does the pulse related to the inverted phase travel to the right?

[Rodrigo Schmidt]

So I'm having a introductory study on waves and there's something that i can't understand when dealing with reflections on a fixed end. We have the general solution for the wave equation:
$y(x,t)=f(x-vt)+g(x+vt)$
Supposing that the fixed point is in the origin we have the boundary condition:
$y(0,t)=0$
Which leads to:
$f(-vt)=-g(vt)$
Using this, for an arbitrary pulse going to the left we have the solution (While this may seem strange, the equation below fits the boundary conditions and the wave equation):
$y(x,t)=g(x+vt)-g(vt-x)$
Therefore, as far as I'm concerned, before the reflection, the pulse would be, normally, traveling to the left and there would be something like a "virtual pulse" with inverted phase and position travelling, too, to the left. That would mean that when the real pulse reaches the origin, there would be no reflection. There's obviously an error in my logic and the book conclusion on the acquired solution is that the initially virtual pulse is traveling to the right (Which leads to the correct conclusion). What I'm not seeing on this second term? What leads to the conclusion that the pulse related to it travels to the right?

 

[Orodruin]

What leads to the conclusion that the pulse related to it travels to the right?

The phase moves to the right.

In other words, after time vt it will have the same value at position x as it had at position x-vt at time t=0.