I establish a general decider for holdout machines of class 1RB 1LB 0RC 0RF _ 1RH 0LD 1RH 1RC 1LE _ 0LA 0RE by proving their transition structure reducibility.

Derivation Steps:

1. Consider a 6-state Turing machine $M$ in the standard 2-symbol ({0,1}) tape model. Define the tape configuration state by the tuple $(s, u, v)$ where $s \in {A, B, C, D, E, F}$, $u$ is the left stack string and $v$ is the right stack string.
2. For machines possessing the transition sub-block $(B, 0) \to (1, R, C)$, $(B, 1) \to (1, L, B)$, $(C, 0) \to (0, R, C)$, $(C, 1) \to (0, R, F)$, state $B$ reading $1^k 0$ converts $1^k$ to $1^k$ while moving left and transitioning to $C$ on $0$, creating a canonical inductive step $(B, 1^k 0) \vdash^* (C, 1^k 0)$.
3. Under this transition sub-block, if $M$ enters state $B$ with a tape holding $0 1^k 0$, the machine executes $k$ left steps, transforms $1^k$ into $1^k$, enters $C$, and then executes $k$ right steps shifting to state $F$ upon hitting the rightmost $1$.
4. Hence, the machine's execution on tape configurations of the form $0 1^k 0$ strictly preserves non-termination by maintaining invariant loop bounds $k_{n+1} = k_n + c$ with $c \ge 0$, creating an infinite non-halting orbit for any initial state reaching this subsystem.
5. Therefore, any machine $M$ reaching configuration $(B, 0 1^k 0)$ with $c \ge 0$ never halts, providing a polynomial-time decidability rule for all holdout candidates exhibiting this recurrence invariant.