No node-consistent extension of $(0,-1)$ is symmetric about $1/2$

For a finite word $A$ of integers, an in-order pair $(x,y)$ of $A$ ($x$ occurs before $y$) is open if $2y-x\notin A$ and $3y-2x\notin A$. It lies at the node $(k,\,x\bmod 2^k)$ of the binary residue tree, where $k=\nu_2(y-x)$, and it forces the bit $\mathrm{bit}_k(y)=\lfloor y/2^k\rfloor\bmod 2$ there. The word satisfies SAFE(ii), or node consistency, if no node is forced to both bits by open pairs: any two open pairs at the same node force the same bit. There is no finite word

satisfying SAFE(ii) whose support is invariant under $u\mapsto 1-u$. Invariance would force $2\in A$ and $\min A+\max A=1$, so the extremes would have opposite parity, contradicting the extremal-parity lemma. In the tree-witness approach to Erdős Problem 195 this excludes the natural symmetric (reflection) completion of the even half that arises when releasing the root.

Formalization note: SAFE(ii) is stated inline exactly as the second conjunct of Erdos195.exists_treeLt_safe_iff. Positions are List.idxOf, $\nu_2$ is padicValInt 2, and $\mathrm{bit}_k$ is bitOne from Definitions.Def_Erdos195_defs. Only the closure $u\in A\Rightarrow 1-u\in A$ is assumed. Every tree-safe word satisfies SAFE(ii).