How does openGL come to the formula F_depth and and is this the window viewport transformation

#point no.1

after transforming points via the projection matrix , we end up with the point in the range [-1,1],
but, in the depth testing chapter , the author mentions that
F_depth = 1/z-1/far /(1/near - 1/far) converts the view space coordinates i.e. z=z_eye is transformed from [-1,1] to [0,1] .

I've followed this thread, and one of the members tell me that the formula F_depth is actually a composition for a series of steps done, and outlines this step:

    z_Clip = C*z_eye+D*W_eye
    w_Clip = -z_eye
    where C=-(f+n)/(f-n), D=-2fn/(f-n).
    Projective division:
    z_ndc = z_clip/w_clip
    Depth range:
    depth = a + (a-b)*(z_ndc+1)/2
    where glDepthRange(a,b) .

I tried composing the formula as he had suggested, but this is completely different from the F_depth formula given in learnopenGL.

#point no. 2

Also another member tells me that [-1,1] to [0,1] is the window viewport transformation, which has a different formula itself.

So, all of this is not making any sense to me(having 3 different formulas and explanations for the same thing that also for openGL), I'll bullet the queries I have regarding these contradicting ideas:

Is F_depth a composition of transformations from view space to window space.
Are the depth range transformation and viewport transformation the same? Why do they have different forumulas(one in Point no.1 and another in this link
How is the F_depth formula obtained? Or how is does the composition of transformations done to convert the world space point to [0,1] result in F_depth?

Solution

In the answer of @horxCoder are missing some steps, which I want to clarify.

In the tutorial LearnOpenGL - Depth testing is claimed that the depth at claimed at Perspective projection is

depth = (1/z - 1/n) / (1/f - 1/n)

where z is the distance to the point of view, n is the distance to the near plane and f is the distance to the far plane of the Viewing frustum.

The question is why the depth of a fragment is given by the above formula?

The symmetrically perspective projection matrix is (see OpenGL Projection Matrix):

1/(ta*a)  0      0              0
0         1/ta   0              0
0         0     -(f+n)/(f-n)   -2fn/(f-n)
0         0     -1              0

For the depth we are just interested in the z and w component. For an input vertex (x_eye, y_eye, z_eye, w_eye), the clip space z_clip and w_clip components are computed by:

z_Clip = C * z_eye + D * W_eye
w_Clip = -z_eye

where

C = -(f+n) / (f-n)
D = -2fn / (f-n)

The normalized device space z coordinate is computed by a Perspective divide

z_ndc = z_clip / w_clip

The normalized device space z coordinate is mapped to the depth range [a, b] (see glDepthRange):

depth = a + (a-b) * (z_ndc+1)/2

When we assume tha the depth range is [0, 1] and the input vertex is a Cartesian coordinate (x_eye, y_eye, z_eye, 1), this leads to the following:

             z_eye * -(f+n) / (f-n)  +  -2fn / (f-n)
depth  =  (------------------------------------------ + 1) / 2
                          -z_eye

And can be transformed

             -z_eye * (f+n)  -  2fn
depth  =  (-------------------------- + 1) / 2
               -z_eye  *  (f-n)

             -z_eye * (f+n)  -  2fn  +  -z_eye * (f-n)
depth  =  ---------------------------------------------
                      -z_eye  *  (f-n)  * 2

             -z_eye * (f+n+f-n)  -  2fn
depth  =  -------------------------------
               -z_eye  *  (f-n)  * 2

             -z_eye * f  -  fn           -f (n + z_eye)  
depth  =  -----------------------   =   ----------------
               -z_eye * (f-n)             z_eye (n - f)

Since the view space z axis points out of the viewport, the z distance from the point of view to the vertex is z = -z_eye. This leads to:

          f (n - z)        1/z - 1/n
depth  =  -----------  =  -----------
          z (n - f)        1/f - 1/n