Manuscript En08-B@1 · En08-B.tex · lemma · 23cac52a

Lemma: Finite complete-channel telescoping

The connecting theorem › A complete-channel error lemma · lines 802–830 · proof included or linked

En08-B@1#lem:telescoping, En08-B@1#eq:telescoping

Let $E, D, G_{1}, \dots, G_{n}$ be compatible channels, and let their implemented counterparts be compatible channels on the same comparison interfaces. Then

d_{⋄} (D G_{n} \dots G_{1} E, D G_{n} \dots G_{1} E) \leq d_{⋄} (E, E) + k = 1 \sum^{n} d_{⋄} (G_{k}, G_{k}) + d_{⋄} (D, D) .

The left-hand side is also at most one.

Proof. Subtract the two compositions and replace one factor at a time. Each term contains exactly one difference of corresponding factors, with channels before and after it. Diamond-norm contraction under those compositions bounds that term by the diamond norm of the factor difference. The triangle inequality and division by two give (eq:telescoping). The distance between two channels in half-diamond normalization is at most one.

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En08-B@1#eq:telescoping · line 807
$d_{⋄} (D G_{n} \dots G_{1} E, D G_{n} \dots G_{1} E) \leq d_{⋄} (E, E) + k = 1 \sum^{n} d_{⋄} (G_{k}, G_{k}) + d_{⋄} (D, D) .$

Lemma: Finite complete-channel telescoping

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