Certified clipping returns the exact convex polygon overlap

Clearing one common positive denominator from the three rational points in an orientation test multiplies the determinant by a positive square, so its sign is unchanged and all scaled coordinates are integers. Therefore [F2] returns the exact side classification, including boundary zero, for every clipping pass.

The algorithm's finite certificate accepts only a simple counterclockwise cycle whose vertices lie in the closed left half-plane of every directed edge, with any collinear consecutive triple ordered monotonically. Each edge is therefore a supporting boundary segment of the convex hull of the listed vertices; the simple closed cycle with positive area traverses that hull boundary once, so the certified input is the stated convex polygon. Now fix one directed edge of $Q$ and write $s(X)$ for its signed edge determinant. If $s(S),s(E)\ge 0$, the whole segment remains and the traversal appends $E$. If $s(S)\ge 0>s(E)$, it appends the crossing point; if $s(S)<0\le s(E)$, it appends the crossing point and $E$; and if both signs are negative, it appends nothing. In either crossing case, $s(S)-s(E)$ has the strict sign of the crossing and is nonzero, and $t=s(S)\mathrm{/}(s(S)-s(E))$ lies in $[0,1]$. If an endpoint lies on the clipping line, the formula gives that endpoint exactly. Thus each pass returns the cyclic boundary of the preceding convex intersection with this closed half-plane, preserving isolated boundary vertices and degenerate point or segment outputs.

Induction over the finite edge list of $Q$ shows the final polygon is $P$ intersected with every defining half-plane of $Q$, hence is exactly $P\cap Q$. Each pass processes a finite list and adds at most two crossings per clip edge, so the finitely many passes terminate.

The final vertices are rational because every crossing uses rational operations with a nonzero exact denominator. For a nondegenerate simple polygon the shoelace sum equals its signed area; the output is counterclockwise, so this is its nonnegative Lebesgue area. A convex intersection with fewer than three noncollinear vertices is empty or lies in an affine line. By [F4], the line and all its subsets are Lebesgue null, so the returned area is zero in these degenerate cases.