i know..
i meant i read through the pages, it seemed that some people got the idea that the wheel that provide forward thrust to the plane.
which is not.

No it doesn't. It basically cancels out all the friction that is supposed to pull the plane backwards, well, because it has the ability to roll. That's the gist of it. Spin the conveyor belt and the wheel will simply roll, either forwards or backwards. It doesn't provide any thrust at all.

Or put it this way: The plane isn't advantaged in any way if you got the conveyor belt to spin forwards either.

Heck let me put in the force analysis:
Suppose the conveyor belt's already moving backwards, with the plane also moving backwards. Make some simplifying assumptions:

1) The wings do not generate any 'negative lift' if the plane move backwards
2) Fuel isn't lost

These 2 assumptions will be needed to ensure the weight of the plane on whatever object it rests upon stays constant.

Right, from A level/SAM (I think)/F6 Physics, we model friction by F = mu*N, mu the coefficient of friction, N the normal force on the belt

So if the thrust is less than this frictional coefficient (we assume for simplicity's sake that the coefficient of sliding friction = coefficient of static friction = mu. Analysis still holds because usually coefficient of sliding friction < coefficient of static friction)

Suppose the thrust is larger than the coefficient of friction. The plane will accelerate relative to this belt, and the key word is, it will continue accelerating relative to the belt no matter how fast the belt is moving, it only depends on the initial velocity the belt was travelling at when this thrust was applied (remember, I assumed that plane travels backwards with belt). Plane takes off nonetheless because it'll continue accelerating and at some point its velocity to some observer in the airport will turn positive, and take off occurs.

Of course we never see this, because the landing gear would probably shatter into pieces if we attempted it, and even that's assuming we have those mystical super powerful engines that can provide that much thrust.

So, since many of us here agree that the plane can take-off even when it is on the conveyor belt, can we use this to decrease the length of the runway?

*Let's assume that we can manufacture a conveyor belt that is strong enough

No, the thrust of the plane has absolutely nothing to do with the speed of the conveyor belt. The thrust will only determine the acceleration of the plane relative to the conveyor belt supposing the force by the engines is larger than the frictional force. So the question to ask is whether the plane will accelerate relative to the belt. The speed of the conveyor belt becomes irrelevant in that context, and that's the point.
In the case of the 'free spinning wheel', basically even without thrust, and in the presence of a moving conveyor belt, ideally the plane will not move, even though there is friction between the tyre and the belt; this is because the conveyor belt simply causes the wheel to roll, and doesn't move the plane at all. The wheel will continue rolling and rolling matching the speed of that conveyor belt, even with 0 thrust. A simple experiment (I used thumbscrews - which are far from ideal, if possible get as round a cylinder as possible with moderate mass (basically not too light)) is to take a cylinder and put it under say a plastic ruler or a flat surface with enough friction. Pull that ruler and you'll see that the object will just roll and stay at the same place. Why? Conservation of momentum of that object. It'll also seem to roll slowly in the direction that you pulled the ruler. Why? Rolling resistance that I discussed earlier. But in the ideal case, the rolling resistance doesn't exist, and thus the plane will stay in the same place no matter how quickly you move the belt. The semi ideal case can be modelled by adding an extra 'dragging force' term, the rolling resistance. If the thrust can handle that, we're done, and plane will take off. That's it.