A.9:5 - Archetypal Grounding
These six cases are constructed model checks, not empirical performance reports. Read each row from the intended result through its model to the warranted answer.
| Intended result and stated model | Selected law, result and limit |
|---|---|
| Total of disjoint exact resource contributions 2 and 3 on one additive quantity scale | Addition gives 2 + 3 = 5 under B.1.6. No weakest-part cap applies to this total; exceeding either part creates no new whole. |
| Probability that both independent components succeed, with probabilities .9 and .8 | The joint-success model selects multiplication: .9 × .8 = .72. A minimum would not calculate that probability. |
| Probability that at least one of those same independent components succeeds | The alternative-success model selects 1 - (1-.9)(1-.8) = .98. Identical marginals and unchanged components support a different law because the intended event changed. |
Compose f(x)=x+1 and g(x)=2x in the required order | g(f(x))=2x+2 differs from f(g(x))=2x+1. Keep the order; no Characteristic or measurement Scale is required for this function-composition claim. |
| Reorder or repartition disjoint exact additive resource contributions | Exact addition permits the stated reorder and regrouping. That argument establishes neither arbitrary floating-point regrouping nor independence of repeated-source evidence. Check the actual numerical and source conditions when they matter. |
| New evidence defeats independence in the reliability cases, while the assembly retains its identity | Withdraw the .72/.98 computations that used independence. Obtain a justified dependence model or return only a supported bound or the separate marginals. The same assembly remains the subject unless its independent identity rule requires a different conclusion. |
For the additive construction, the singleton base case returns 2 from the one-element input [2]. It does not claim 2 + 2 = 2. Nor does writing the same observation twice create two independent observations.