Four of the corners of a cube can be cut off by taking its intersection with a tetrahedron

$\mathrm{tetra}≔\mathrm{PolyhedralSet}\left(2\left[\left[1\,1\,1\right]\,\left[1\,-1\,-1\right]\,\left[-1\,1\,-1\right]\,\left[-1\,-1\,1\right]\right]\,\left[x\,y\,z\right]\right)\:$$\mathrm{cube}≔\mathrm{ExampleSets}:-\mathrm{Cube}\left(\left[x\,y\,z\right]\right)\:$$\mathrm{t\_c\_intersect}≔\mathrm{tetra}\mathbf{intersect}\mathrm{cube}$

$\mathrm{t\_c\_intersect}≔\{\begin{array}{lll}\mathrm{Coordinates}& \:& \left[x\,y\,z\right]\\ \mathrm{Relations}& \:& \left[-z\le 1\,z\le 1\,-y\le 1\,y\le 1\,-y-z-x\le 2\,-x\le 1\,y+z-x\le 2\,x-y+z\le 2\,x\le 1\,x+y-z\le 2\right]\end{array}$

$\mathrm{Plot}\left(\mathrm{t\_c\_intersect}\right)$

Construct a tetrahedron and a cube

$\mathrm{tetra}≔\mathrm{ExampleSets}:-\mathrm{Tetrahedron}\left(\right)$

$\mathrm{tetra}≔\{\begin{array}{lll}\mathrm{Coordinates}& \:& \left[x_1\,x_2\,x_3\right]\\ \mathrm{Relations}& \:& \left[-x_1-x_2-x_3\le 1\,-x_1+x_2+x_3\le 1\,x_1-x_2+x_3\le 1\,x_1+x_2-x_3\le 1\right]\end{array}$

$\mathrm{cube}≔\mathrm{ExampleSets}:-\mathrm{Cube}\left(\right)$

$\mathrm{cube}≔\{\begin{array}{lll}\mathrm{Coordinates}& \:& \left[x_1\,x_2\,x_3\right]\\ \mathrm{Relations}& \:& \left[-x_3\le 1\,x_3\le 1\,-x_2\le 1\,x_2\le 1\,-x_1\le 1\,x_1\le 1\right]\end{array}$

The tetrahedron tetra is a subset of the cube cube

$\mathrm{tetra}\mathbf{subset}\mathrm{cube}$

$\mathrm{true}$

But cube isn't a subset of tetra

$\mathrm{cube}\mathbf{subset}\mathrm{tetra}$

$\mathrm{false}$

Any point in a set will return true when tested with in

$c≔\mathrm{ExampleSets}:-\mathrm{Cube}\left(\right)$

$c≔\{\begin{array}{lll}\mathrm{Coordinates}& \:& \left[x_1\,x_2\,x_3\right]\\ \mathrm{Relations}& \:& \left[-x_3\le 1\,x_3\le 1\,-x_2\le 1\,x_2\le 1\,-x_1\le 1\,x_1\le 1\right]\end{array}$

$\left[0\,0\,0\right]\mathbf{in}c$

$\mathrm{true}$

To find the face on which the point resides, see PolyhedralSets[LocatePoint]