Lemma 51 ([Clearance from the original boundary]). Let S be an independent full walk from v , stopped only on the original boundary at τ_D , and let γ be the completed interface. Put d=dist(v,∂D) and M= S_j:j<τ_D, S_j is adjacent to γ . …

We adapt the full-walk argument of . We first establish pure-arc repulsion. For y∈∂_+ , choose any deterministic lower bound b≤d_D(y,∂_-) . At successive fixed-ratio scales between b and r , take the component of B(y,ρ)∩D accessible from y , and its circular crosscut separating y from ∂_- . The clearance construction used in Lemma 50 ([No additional sampled excursions]) supplies a barrier attached to the two positive boundary pieces on either side of y . Its endpoint pockets stay on ∂_+ , since their diameter is smaller than b . Explore the interface only up to the next outer crosscut. Its exact sign transcript leaves the inner barrier compatible, so Proposition 45 ([Uniform barriers]) gives a fixed conditional chance of avoiding this crosscut and every smaller access to y . Iterating, then reflecting for y∈∂_- , gives constants C,ζ>0 such that d_D(y,γ)<r ≤C(r/b)^ζ+o_d(1), when r,b are fixed positive multiples of d and r<b . The lattice error is uniform for these fixed multiples. This is the deterministic crosscut argument of , with its field input replaced by our exterior version of Proposition 45 ([Uniform barriers]). Fix r=ηd , and condition on γ . At the first walk visit to M at distance less than r from ∂D , the strong Markov property and the planar walk estimate give a fixed probability of exiting D in the ball of radius 2r+O(1) about a closest boundary point. One can force the walk to surround that point inside this ball, which forces the exit. Its continuation and the adjacent interface edge then give an internal path of diameter at most 4r+O(1) from the exit Y=S_ τ_D to γ . Consequently dist(M,∂D)<r ≤Cd_D(Y,γ)<4r+O(1). Here Y , unlike the interface landing, is independent of γ . Conditional on Y , apply (GB2) whenever its internal distance to the opposite arc exceeds s=√η\,d , using the deterministic lower bound b=s in that estimate. This gives Cη^ ζ/2 +o_d(1) . It remains to bound exits within internal distance s of both sign arcs. Delete all paths of diameter at most 2s joining the two arcs and retain the component of v . Its new boundary lies in two balls of radius 4s . To verify this elementary fact, classify each deleted crosscut by which marked sign-change endpoint it separates from v . Any two crosscuts in the same class which touch the retained component intersect: otherwise their separation order prevents one of them from touching that component. Their diameter bound puts the whole class in one ball. There are two classes. An exit near both arcs is beyond this new boundary. The walk hitting estimate for these two balls gives C(s/d)^ 1/2 =Cη^ 1/4 . Combining with (GB3) proves (GB1). Every nonexterior old-prefix landing is a vertex of M , pathwise, because the prefix is contained in γ and the full walk has not yet exited D . This proves the simultaneous assertion without conditioning on any landing event or on any stopping time.