absolute_floor_unique_normalized_01
plain-language theorem explainer
On any absolute-floor configuration carrier, two normalized two-point recognition floors that share the same marked point must use the same Boolean orientation and the same pointwise recognition-work cost. Foundation authors cite this when showing the Boolean floor is unique once empty/marked orientation and unit cost are fixed. The proof is a one-line application of the normalized two-point uniqueness lemma.
Claim. Let $Config$ be a nonempty configuration space admitting an absolute-floor witness. Fix a marked point $m \in Config$. If $(C_1,\phi_1)$ and $(C_2,\phi_2)$ are two normalized two-point recognition floors on $Config$ with mark $m$ (each a unit-normalized recognition-work cost plus an equivalence $Config \simeq \mathrm{Bool}$), then $\phi_1 = \phi_2$ and $C_1(\Gamma) = C_2(\Gamma)$ for every configuration $\Gamma$.
background
The Unified Forcing Chain module proves that T0–T8 are forced from the cost foundation (Recognition Composition Law with normalization $F(1)=0$ and calibration $F''(1)=1$). The chain opens at T-1, the absolute floor: a meta-language that can distinguish propositions, plus a non-singleton universe of discourse.
A configuration space supplies an empty configuration, a binary join, a consistency predicate, and an independence relation. An absolute-floor witness records bare distinguishability (some propositions differ) and a nontrivial specification on the carrier. A normalized two-point recognition floor is the abstract Boolean picture without smuggling $\mathrm{Bool}$ in as an axiom: one empty/consistent point, one marked inconsistent point, unit recognition work on that marked inconsistency, exhaustive two-point support, and an equivalence showing $\mathrm{Bool}$ is only the canonical representative.
The surrounding comment stresses that bare T-1 distinguishability does not by itself name false/true or fix a unit cost scale; normalization is the explicit orientation step.
proof idea
One-line term wrapper. Apply the upstream lemma that any two normalized two-point recognition floors with a shared mark have identical Bool equivalences and pointwise-equal recognition-work costs. The absolute-floor witness is only ambient context (the carrier is a legitimate floor); it is not inspected in the term. All algebraic content lives in that uniqueness lemma and in the structure fields (mark differs from empty, exhaustive two-point cover, consistency equals emptiness, unit cost on the mark).
why it matters
This uniqueness certificate closes the normalization step at T-1 in the complete inevitability chain. Bare absolute-floor distinguishability does not name false/true or fix unit recognition work; the normalized two-point floor supplies that orientation, and uniqueness shows any two such normalizations with the same mark coincide. The Boolean floor is therefore canonical rather than a modeling choice.
That underwrites the forcing narrative above T-1: T0 (logic from cost minimization, consistency is cheap) sits on a uniquely determined two-point recognition floor, before MP, discreteness, ledger symmetry $J(x)=J(1/x)$, unique $J$, $\varphi$, the eight-tick octave, and $D=3$. No downstream dependents are recorded yet; the result is infrastructure for the absolute-floor layer of the chain.
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