all_r0_assignments_count
plain-language theorem explainer
There are exactly 24 raw sector-to-value assignments of the four candidate $r_0$ yardstick values, before structural filtering. Anyone checking the O1 yardstick choice-set enumeration cites this count as the size of the unfiltered permutation pool. The proof is a one-line `native_decide` on the concrete finite list.
Claim. The finite list of all $r_0$ sector assignments obtained by taking every permutation of the four-element candidate $r_0$ value pool (and retaining those that parse as well-formed assignments) has length $24$.
background
The module treats the O1 yardstick discussion as a finite combinatorial search. One starts from four candidate $B_{\mathrm{pow}}$ values and four candidate $r_0$ values, forms every sector-to-value assignment by permutation, then filters by the structural constraints used in the yardstick principle.
Here the object counted is the unfiltered $r_0$ side: allR0Assignments is the list of every permutation of the $r_0$ value pool that successfully converts to an R0Assignment. The companion length notion is ordinary finite-list length (the same Nat length used for traces elsewhere in the foundation). Downstream filters impose sum targets such as $V\cdot W+E_{\mathrm{passive}}=147$ and related structural constraints; this theorem only sizes the raw pool.
proof idea
One-line wrapper: native_decide evaluates the concrete closed term allR0Assignments.length and checks equality with $24$. No lemmas are invoked beyond the definitional unfolding of the permutation list and the Nat length function.
why it matters
In the Recognition yardstick story, particle masses sit on a $\varphi$-ladder with a sector-dependent yardstick; assigning the four $r_0$ candidates to sectors is part of fixing that ladder. This count anchors the O1 choice-set enumeration: the raw $r_0$ assignment space is $4!=24$ before structural cuts. The module claims that after those cuts the valid $r_0$ (and $B_{\mathrm{pow}}$) choice sets collapse to singletons, so the enumeration is the explicit finite witness behind uniqueness of the yardstick assignment. No downstream theorem currently depends on this lemma in the graph; it is a verification checkpoint for the combinatorial side of the mass-yardstick story rather than a step in the T0–T8 forcing chain.
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