Flip a cell of the four-part partition exactly when it contains an edge $uv$ of $B$ with no walk of length at most $6$ in $B̄$. Then every edge of $B'$ that is an edge of $B$ has a short walk in $B̄$, by the choice of the flipped cells. Every edge of $B'$ that is a non-edge of $B$ lies in a flipped cell, and comparing it with the edge that caused the flip gives a walk of length at most $6$ in $B$ between its ends, by a case analysis over the adjacencies to the representatives. Walks between the two sides of a bipartite graph have odd length, so the bound $6$ improves to $5$.

The construction is adapted from an earlier formalization that proves the statement with bound $6$ for flips given by Boolean toggles, see PROVENANCE.md.