Proposition: Certificate-to-channel error bridge

Suppose the analytic table enclosures obey (eq:certificateerror) and all implemented operational factors, on the same complete interfaces, have summed half-diamond allowance $\eta$. Then the implementation $\tilde{\Gamma }$ satisfies

The bound includes arbitrary untouched references. It compares installed finite channels, not continuum sharp-product kernels.

Proof. The table comparison is (eq:enclosurechannel). Exact classical copying and payload transport are common postprocessing, so they cannot increase this distance. For the remaining implementation differences, replace one compatible channel factor at a time. Pre- and postcomposition by channels contract diamond norm. Summing those differences gives $\eta$; the distance between channels is at most one Watrous.