# Le Chatelier's principle with pure solids and liquids

In the following reaction:

$$\ce{A(s) <=> B(g) + C(g)}$$

since adding a pure solid is not supposed to shift the equilibrium in any way, does that mean that I can essentially remove all of the solid except for 1 molecule, and the equilibrium will not shift at all since the concentration of the solid stays the same? Similarly, does this mean that I can essentially fill up the container with the solid, and no extra B and C will be produced?