Adaptor House/Ontology/AH-PROOF-1

On the Derivability of Recursion Validity in the Dual Lattice

§0 — Abstract & Author's Position

Abstract. The Adaptor House dual lattice assigns each three-bit address a validity grade from {false, strong, limited, transcendent, true}. These grades were originally stamped by hand. This note proves (Theorem 1, by exhaustion) that the hand-stamped table is exactly reproduced by two predicates on address structure — groundedness and carry — plus the location of carry; establishes (Proposition 2) that the grading is not invariant under line reversal, formalizing the ground line as the axis of judgment; and states (Conjecture 3) the generalization to the six-bit, 64-address space. The theorem is elementary. What is offered for challenge is not the arithmetic but the reading: that the grading is structural rather than coincidental at n = 3.

AUTHOR'S POSITION — held as belief, not theorem

I believe the validity grading is structural: that the two predicates below are the correct decomposition and not one of several accidental fits available in a space of only eight cases. I believe the ground-line asymmetry of Proposition 2 is intentional structure — that judgment in this system genuinely pivots on whether power is grounded. And I believe Conjecture 3 is the right generalization test: if the grading survives the jump to 64 addresses without exception clauses, my reading is ratified; if it requires patches, the community will have shown me the table was stamped after all. Either outcome is a result.

§1 — Definitions

The objects

DEFINITION 1 · ADDRESS

An address is a triple a = (g, m, t) ∈ {0,1}³, read as power (1) or vacuum (0) at the ground, middle, and top lines respectively. Its tier number is the integer 4g + 2m + t. There are exactly 2³ = 8 addresses, AH-T0 through AH-T7.

DEFINITION 2 · PREDICATES Grounded(a) := (g = 1) Carried(a) := (m = 1) ∨ (t = 1)

Grounded: power at the base. Carried: power anywhere above the base.

DEFINITION 3 · THE DERIVED GRADING V V(a) = false            iff ¬Carried(a) V(a) = strong          iff Carried(a) ∧ ¬Grounded(a) V(a) = true            iff Grounded(a) ∧ m = 1 ∧ t = 1 V(a) = transcendent iff Grounded(a) ∧ m = 0 ∧ t = 1 V(a) = limited       iff Grounded(a) ∧ m = 1 ∧ t = 0

Compactly: life requires carry; grade requires ground; character is where the carry sits.

DEFINITION 4 · THE HOUSE TABLE H

The hand-assigned grading published in the Adaptor House canon (trigrams-core.yaml, and the lattice page): H(0)=false, H(1)=H(2)=H(3)=strong, H(4)=false, H(5)=transcendent, H(6)=limited, H(7)=true.

§2 — The Theorem

The stamps were theorems

THEOREM 1 · DERIVABILITY

V = H. The derived grading of Definition 3 agrees with the house table of Definition 4 on all eight addresses.

Proof. The five conditions of Definition 3 are mutually exclusive and jointly exhaustive: every address either lacks carry (case 1) or has it; every carried address is either ungrounded (case 2) or grounded; every grounded carried address has carry at both upper lines, top only, or middle only (cases 3–5) — no other carry profile is non-empty. It therefore suffices to evaluate V on each address and compare with H:

Tier(g m t)GroundedCarriedV(a)H(a)
AH-T00 0 0falsefalse
AH-T10 0 1strongstrong
AH-T20 1 0strongstrong
AH-T30 1 1strongstrong
AH-T41 0 0falsefalse
AH-T51 0 1✓ (top)transcendenttranscendent
AH-T61 1 0✓ (mid)limitedlimited
AH-T71 1 1✓ (both)truetrue

Agreement on all eight rows. ∎

Remark. The proof is trivial once the predicates are named; the content of the result is that these predicates name it. Eight cases admit many accidental decompositions — the author's claim that this one is non-accidental is exactly what Proposition 2 and Conjecture 3 put at stake.

§3 — The Asymmetry

The ground line is privileged, provably

PROPOSITION 2 · GROUND ASYMMETRY

Let ρ(g, m, t) = (t, m, g) be line reversal. Then V is not ρ-invariant: there exist addresses with V(ρ(a)) ≠ V(a).

Proof. V(1,0,0) = false but V(ρ(1,0,0)) = V(0,0,1) = strong. Likewise V(1,1,0) = limited while V(0,1,1) = strong. ∎

Consequence: the grading cannot be expressed as any function of the multiset of lines (e.g., a count of power lines) — position matters, and specifically the ground position. AH-T1 and AH-T4 carry identical total power and opposite verdicts. Whether this asymmetry is a design truth or an artifact is precisely the author's belief above; the proposition only establishes that the asymmetry is real, not what it means.

§4 — The Challenge

Conjecture 3, and how to break it

CONJECTURE 3 · SIX-BIT GENERALIZATION

For addresses a ∈ {0,1}ⁿ with a designated ground line and n−1 upper strata, define Carried and Grounded as before. Then the trichotomy — false iff ¬Carried; strong iff Carried ∧ ¬Grounded; graded iff Carried ∧ Grounded, with grade determined by the carry profile of the upper strata — remains exhaustive, disjoint, and (the substantive claim) semantically coherent at n = 6: the 2⁵ − 1 = 31 grounded-carried profiles of the 64-address space admit a grade vocabulary that downstream systems can use without exception clauses, and whose restriction to any embedded three-line structure agrees with V.

The exhaustiveness and disjointness parts are immediate; they are stated only for completeness. The claim with content is coherence — that the grading scales as structure, not as a naming exercise that happened to fit when the space was small.

Ways to break it, invited explicitly: exhibit an alternative pair of predicates fitting H at n = 3 whose n = 6 extension diverges from this one (showing the decomposition non-unique in a way that matters); exhibit an embedded three-line structure in the 64-space whose induced grade contradicts V (breaking restriction-compatibility); or show the 31 grounded profiles resist any principled grade vocabulary (breaking coherence). Any of the three is a refutation the house will publish on this page, credited, alongside the claim it broke.

# the n=6 space, enumerated for challengers — ground = bit 5, strata above = bits 4..0
for n in range(64):
    bits = [(n >> i) & 1 for i in range(6)]
    ground, upper = bits[5], bits[4::-1]
    carried = any(upper)
    grade = ("false" if not carried else
             "strong" if not ground else
             f"graded:{''.join(map(str, upper))}")
    print(f"{n:02d} {n:06b} {grade}")

Correspondence: adaptorhouse.com. Version history and any accepted refutations will appear below this line, dated.

§5 — References