Proposition: No Landauer cost from channel equality alone

Equality of observable channels, including equality with partial trace or a reset channel, does not determine a unique heat, work, or cyclic reusable-memory reset cost.

Proof. Let $M$ be a qubit memory and $A$ an independent blank qubit in $|0⟩$. A swap gives, including arbitrary references and after regrouping,

The observable memory channel is always the same reset channel. Nevertheless, for every $ϵ>0$ we may choose

The swap conserves $H_M+H_A$, but if the initial excited-state probability is $p$, then

Thus the observable channel fixes neither these energy transfers nor the cost of restoring the consumed blank. The input diary has moved to $A$.

Even after the memory Hamiltonians have been fixed, an independent unobserved process can change the total implementation ledger without changing the memory channel. For example, choose a fuel qubit $F$ initially excited and a designated thermal reservoir qubit $H$, both with gap $ϵ$. At inverse temperature $\beta =(k_{\mathrm{B}}T{)}^{-1}$, the latter has excitation probability

Swapping $F$ and $H$ increases the designated reservoir energy by $ϵ(1-r)$ while consuming fuel. It acts independently of $QM$. Repeating such independent spectator processes changes the designated reservoir energy increase while leaving the observed reset channel unchanged.

These are not cyclic erasure protocols unless the blanks, fuel, and controls are also restored. They prove underdetermination, not a violation of a thermal erasure inequality.