Proposition: Thermodynamic underdetermination

The visible channel $\Phi$, the destructive record channel, and their complete reference identities do not determine a unique physical energy transfer, bath heat, work cost, or cyclic memory-reset cost.

Proof. In the dilation (eq:dilation), assign the environment Hamiltonian

The reduced channel on the data is unchanged as $ϵ$ varies. If the pre-relaxation data state has antiqubit population $p_1$ in value $1$, the environment's energy increase is

Hence the visible channel alone fixes neither that energy transfer nor the Hamiltonian needed to interpret it. A pure blank environment has not thereby been identified as a thermal reservoir.

Even after chosen Hamiltonians are fixed, an independent unobserved process can alter a designated reservoir ledger without altering the data or record channel. For example, take a fuel qubit initially excited and a Gibbs reservoir qubit, both with gap $ϵ$. At inverse temperature $\beta$, the reservoir excitation probability is

Swapping the two qubits increases the designated reservoir energy by $ϵ(1-r)$, while acting trivially on the visible interface. Repeating this independent process changes that ledger without changing the visible channel.

These examples consume blanks or fuel. They are not cyclic erasure protocols unless those resources and controls are also restored. They establish underdetermination, not a violation of a thermal inequality.