C.29.1:4.5 - Derive the consequence in the strength that survives

Derive the receiving consequence from the comparison.

For an exact calculation, substitute the represented inputs and derive the receiving output. When returning an answer, express it in the quantity originally asked for. For a transported derivation, identify which premises and inference steps are covered. For example, an identity built from compositions of the compared operations gives the corresponding receiving identity on represented inputs. If a proof also uses order, division or an existence premise, establish how that contribution applies in the receiving account. A quantified claim depends on the cases over which it ranges.

A map that combines source cases can erase differences; equality of their images does not establish equality of the original cases. A receiving domain can contain cases outside the map’s image; a result proved only for represented cases does not cover those additional cases. Use an inverse, a separate argument, or a narrower conclusion when the receiving claim requires it.

When the exact queried value is not determined, construct the values compatible with the retained information. For a represented value y, these arise from source cases satisfying F(x) = y and the stated premises. Prove a lower bound L(y), an upper bound U(y), or another relation that holds for all those cases. The resulting interval can answer a threshold question even when it cannot identify one value.

For a decision q ≤ b:

• an upper bound U(y) ≤ b establishes the decision for every compatible case;
• a lower bound L(y) > b rules it out for every compatible case;
• a bound spanning b leaves the decision unresolved.

These are consequences of the bound, so retain its domain, units and relevant time or parameter range. A bound for a sampled instant does not by itself answer what happened between samples.

A relaxation offers another useful transfer. Let S be the allowed source solutions, let T be receiving solutions, and suppose every solution in S has an image in T with the same cost. If the receiving account permits additional solutions, minimizing there can give a lower bound on the source minimum. When the minima exist:

min over T of receiving cost ≤ min over S of source cost.

The reason is that the receiving search includes a cost-preserving image of every source option. Maximization gives an upper bound under the analogous assumptions. An optimal receiving solution that has no source counterpart still supports the bound; it does not supply a feasible source plan. A returned plan requires an allowed source witness.

Where the comparison is approximate, derive the error relation for the needed operation and propagate it through later steps. For example, if a later scalar operation h satisfies |h(a) − h(b)| ≤ K|a − b|, with K ≥ 0, on the relevant interval, an input error at most ε contributes at most Kε at its output. Any additional error introduced by computing h is added to that contribution. Merely adding the errors of successive steps without accounting for amplification can understate the final error.