Reduced two-disk sums of least fundamental disks have no closed exchanges

At least one original boundary is null-homotopic. Otherwise both $X,Y$ are essential normal disks, so their weights are at least $w(D)$. Their weights add to $2w(D)$ and hence both equal $w(D)$. Their vectors add to $2d$; unless both equal $d$, the first coordinate at which they differ from $d$ is smaller than $d$ in one vector and larger in the other. This contradicts lexicographic minimality. Equality of both vectors to $d$ contradicts the assumed nonproportionality. This proves the claimed boundary assertion.

Suppose there is an original double circle. An innermost such circle on the original disk $X$ cuts off an interior disk with no other double circle. It contains no proper double arc: the endpoints of such an arc lie on $\mathrm{\partial }X$, outside this cap, so entry would require crossing its frontier, contrary to the disjoint double-curve position. It is therefore an original interior disk patch. Choose smallest positive weight $m$ among all such patches on both families, using [F3]. For each chosen minimum patch $E$, its annular exchange has an opposite disk cap $B(E)$ in the full resolution; its disjointness, adjacency and weight $m$ follow from the same-copy or between-copy suppliers in [F1]. No proper trace enters $B(E)$ either: its boundary is a circle in the interior of one resolved disk, all proper trace endpoints lie on that disk's original boundary, and traces are disjoint.

Thus every cap in the minimum-leaf construction [F1] is cut only by circles. To specify the application, cut $B(E)$ into planar original patches, root its adjacency tree at the outer frontier, and note that every non-root leaf is an interior original disk patch of weight at least $m$. Total weight $m$ forces a unique such leaf $L(E)$, of weight $m$, and a zero-edge annular corridor from it to the outer frontier unless the cap is that leaf. Iteration gives a finite cycle $E_i\mapsto E_{i+1}\subset B_i$. On each resolved disk the caps are disjoint or nested; uniqueness of their minimum leaf excludes nesting, so the $B_i$ are disjoint. If all corridors are nonempty, the local four-ray check in [F1] applies unchanged: leave the selected circles unswitched, keep every other curve, including every proper arc, regular, and glue each corridor to the preceding one by the original diagonals. No proper trace meets a corridor. The resulting nonempty annular union is closed in the parametrizing normalization and has zero edge weight. Every nonempty component of that actual partial resolution contains a genuine normal tetrahedral disk and hence an original edge point, a contradiction. Consequently some $B_i$ equals its minimum leaf, giving a reciprocal two-cycle at one original double circle.

At that circle both opposite caps are original disk patches. Adjacency makes their cap rays belong to different original families. Regularly switching only this circle swaps the patches: each original disk, with one interior disk removed, is an annulus, which the other disk patch caps back to a disk. The interiors of the swapped patches contain no other double curve, so every remaining circle and proper arc still joins the unchanged parts of different families. Both new disks are therefore embedded and normal by [F2], and retain exactly their respective original boundary circles. In particular the null-homotopic boundary from step 1.1 persists. The disk with that boundary cannot have vector proportional to $d$: fundamentalness makes $d$ primitive, a positive proportional integral vector is then $jd$ for an integer $j\ge 1$, and its abstract normal surface is $j$ copies of $D$ by [F2]. Connectedness would force $j=1$ and an essential boundary. Since the two new vectors still sum to $2d$, neither is proportional to $d$. The patch swap is consequently an eligible two-disk diagram with one fewer double curve, contradicting the stipulated minimum. No double circle exists, proving the result.