Proposition: No Landauer cost from channel notation alone

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

The channel data do not specify the physical Hamiltonians, thermal reservoir, initial resource states, controls, or cyclic restoration requirements. Energy conservation for a declared Hamiltonian is a separate assertion.

Proof. A concrete reset dilation makes the underdetermination explicit. Let $M$ be a qubit memory and let $A$ be an independent blank qubit in $|0⟩$. A swap gives, including arbitrary references,

with the factors regrouped. The observable memory channel is the same reset channel for every choice of a common energy gap

The swap conserves $H_M+H_A$, but for input excited-state probability $p$, the energy changes are

Thus even the energy transfers are not fixed by the reset channel. The blank resource is consumed, and the diary has moved to $A$, not disappeared.

One can also append an independent unobserved process without changing any channel on $QM$. For example, take a fuel qubit $F$ initially in $|1⟩$ and a thermal qubit $H$ with the same gap $ϵ$, initial excited-state probability

Swapping $F$ and $H$ conserves their total energy and increases the designated reservoir's energy by $ϵ(1-r)$, while consuming fuel energy. Whether that energy increment is included in a heat ledger depends on the specified implementation and reservoir convention; it is invisible to the memory channel.

Likewise, an independently driven, thermally isolated spectator with equal endpoint Hamiltonians can absorb actuator work while leaving the observed channel unchanged. Hence observable channel equality cannot fix the total implementation ledger.

These augmented implementations are not cyclic erasure protocols unless the consumed blank, fuel, and controls are also restored. They therefore establish underdetermination, not evasion of a thermal erasure bound.