Why can't a perpetual motion polyhedron exist? I've been thinking about polyhedrons, when placed on a table on a certain face, will tip over and keep tipping over infinitely. I'm trying to prove mathematically that such a polyhedron doesn't exist. 
Rules: the polyhedron does not need to be uniform nor convex, but at any point the mass there is non-negative. this means you can have holes and caves and points and fun stuff. 
Right now I'm stuck because I'm not sure what determines if the polyhedron tips over. Either it's 1. The COM of the polyhedron, when projected onto the table, lies outside of the polygon face touching the table, and 2. The gravitational potential energy, mgh, is lowered by tipping over. 
I understand with some basic geometry that 1 implies 2, but condition 2 does not necessarily imply that it will tip over. 
How do I proceed to prove that such a perpetual polyhedron doesn't exist? 
Edit: Thought about it some more and solved my question. 1 and 2 are equivalent in the real world because if the COM was above the base face, then tipping over would require that mgh first increase before finally adopting its final value after tipping over. If the COM is not above the face then there is no necessary activation energy. Therefore, the polyhedron tips over if and only if the energy strictly decreases, which happens if and only if the COM lies outside the face. After this process the potential energy decreases. 
Now assume that the polyhedron spontaneously remains in perpetual motion. Then the energy is strictly decreasing; but the potential energy clearly has a minimum value, so it can't go on forever. 
 A: It will tip over when the COM is outside the convex hull of the support area.  For a polyhedron, the convex hull will be a convex polygon.  The COM can be lowered by pivoting around the nearest edge of the convex hull.
A: The convex hull of the polyhedron will roll on a horizontal plane spontaneously, from being seated on one face to being seated on another, only if (not necessarily if) the centre of mass of the polyhedron is lowered in doing so. Since the convex hull of the polyhedron has only a finite number of faces, there can be only a finite number of such lowerings, and so the process must terminate in a finite number of steps.
Edit: The above answer considers only the static aspects of the problem. If we allow dynamics into the picture (which is realistic), while (unrealistically) assuming zero friction or other resistance, then a polyhedron can indeed rock backwards and forwards, or roll, forever. If we are fully realistic, perpetual motion cannot occur, for reasons of friction alone.
A: I have studied philosophy of physics and dimensions within logic. My answer is dependent on that type of basis.
I find some of the mathematical questions a bit silly in this case. Isn't the movement of a polyhedron a physical problem?
For the sake of imagination, I can imagine cases where mathematically-physically a polyhedron would tip, for example:


*

*Timeless force applied to the object.

*An assumption of continued motion, for example, a vector, which may not require force if no force is implied.

*A sequential operation implicated in the object, as the others seemed to be saying, for example, numeric or categorical inversion, extrusion, etc.

*An assumption that imbalance creates force, which may be a matter of how the program running the object handles imbalances. To me this would seem like an added timeless force or sequence, or some other type of vector.


All in all, it seems to me to call a polyhedron a perpetual motion machine in mathematical space implies mathematical assumptions about the use of vectors.
