Proposition: Exact truncation versus computed-matrix error

Let $W^{[L]}$ be the exact finite-chain isometry and let $\overline{W}$ be a normalized label-symmetric computed isometry on the same identified outputs. Suppose an independent arithmetic certificate establishes

Set

Then the continuum complete channel and the normalized computed complete channel are at distance at most $e_L$.

For the same packet projection, if $\overline{r}$ is the computed isometry's exact erased weight, then

and therefore

If $\overline{\mathcal{D}}$ is the computed model's exact optimal CPTP decoder, then

Proof. The exact continuum-to-chain channel comparison costs ${\zeta }_L(T)$. The arithmetic isometry comparison costs ${\alpha }_L$ by Theorem thm:C1. The triangle inequality gives $e_L$.

Measuring whether the excitation lies in the source or missed modes is a binary effect on the complete output. Its probability difference is at most the channel distance. The label symmetry makes these probabilities $r_N$ and $\overline{r}$ independent of the code input.

Both exact erasure errors equal their erased weight times $c_d$, by Lemma lem:erasure. For the fixed computed decoder, postcomposition contracts the channel comparison; add its computed-model error.