Grid moves generate isotopy of oriented knots

A knot grid is nonempty. Apply [F1] to each $G_i$, obtaining a finite grid-move sequence to a bridge-like grid $B_i$. Smooth $B_i$ as in that supplier to a bridge diagram $D_i$. Its crossing decorations and strand orientation agree with the original grid diagram, and the supplier proves planar isotopy after smoothing. Thus $D_0,D_1$ represent the same oriented knot. The three-mark splits and strict interval commutations in [F1] are among those specified here: the retained exterior row and column partners provide the two markings outside the stabilization block in the Statement. No unlisted shared-endpoint switch has been used.

By [F2] choose a finite sequence of planar isotopies and oriented Reidemeister moves from $D_0$ to $D_1$. Compose successive planar isotopies, and regard each remaining Reidemeister move as an edge between its two decorated planar-isotopy classes. The sequence now traverses finitely many such classes, with the first and last represented by $D_0,D_1$. If there is no Reidemeister edge, these two diagrams are planar isotopic. Otherwise every intermediate representative is a regular diagram and has a bridge representative in its decorated planar-isotopy class by [F5]; choose the perturbation small enough that its crossings and decorations persist. Choose one bridge representative for each occurrence of a class in this finite sequence, keeping $D_0,D_1$ at its ends, and choose a bridge rectification there. The choices are finite; no stronger choice principle is invoked.

Across a Reidemeister edge, the chosen bridge representatives need only be planar isotopic to that edge's original endpoint diagrams. This is exactly the relative-representative interface of [F4], so their chosen grid rectifications are joined by actual grid moves, with every exterior strand included. Across an interval with only planar isotopy, [F3] joins the chosen rectifications. This also treats the sequence with no Reidemeister edge. The endpoint rectifications may be taken to be precisely $B_0,B_1$, because both suppliers apply to arbitrary rectifications. Concatenating the finitely many constituent finite sequences gives a sequence of allowed commutations, stabilizations and destabilizations from $B_0$ to $B_1$.

Prepend the refinement from $G_0$ to $B_0$ and append the reversed refinement from $B_1$ to $G_1$. Reversing a stabilization gives a destabilization and reversing a legal commutation gives a legal commutation, so all operations remain allowed. This produces the required finite sequence from $G_0$ to $G_1$. The proof needs no cyclic permutation, but allowing those additional moves in the Statement does not change the conclusion. The local suppliers cover all marking types and strand orientations, so the sequence preserves the prescribed oriented-knot convention throughout.