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140 changes: 140 additions & 0 deletions equations/thermodynamics.md
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# Thermodynamic Equations

> Pages 19–21 (§173–§175). The energetic cost of computation.

## Landauer Principle

Every irreversible erasure of one bit of information dissipates at least:
```
E_min = k_B · T · ln(2) [binary]
E_min = k_B · T · ln(r) [radix r, general]
```

At room temperature (T = 293 K, k_B = 1.381 × 10⁻²³ J/K):

| Operation | Minimum energy |
|-----------|----------------|
| Binary bit erase | k_B T ln(2) ≈ 2.80 × 10⁻²¹ J |
| Ternary trit erase | k_B T ln(3) ≈ 4.44 × 10⁻²¹ J |

The ratio is exactly ln(3)/ln(2) ≈ 1.585, which also equals the information ratio
(one trit carries log₂(3) ≈ 1.585 bits). Information per joule is identical for
binary and ternary at the Landauer limit.

```
LANDAUER = CONCRETE = 93 [L(19)+A(11)+N(25)+D(13)+A(11)+U(7)+E(3)+R(4) = 93]
```

---

## Radix Efficiency (Equation 13)

```
η(r) = ln(r) / r
```

| Radix | η(r) |
|-------|--------|
| 2 | ≈ 0.347 |
| 3 | ≈ 0.366 ← maximum among integers |
| 4 | ≈ 0.347 |
| 5 | ≈ 0.322 |
| e | = 1/e ≈ 0.368 ← global maximum |

Ternary achieves the maximum radix economy among integer bases because 3 is the
integer closest to e ≈ 2.718. (Proof: see [`../proofs/ternary-efficiency.md`](../proofs/ternary-efficiency.md).)

```
RADIX = GAUSS = TANH = FIELD = 57
```

---

## Reversible Logic Entropy (Equation 14)

For a reversible computation:
```
ΔS_comp ≥ 0, with ΔS_comp → 0 as reversibility → 1
```

The minimum entropy production per gate operation is zero for perfectly reversible gates
(Bennett 1973). In practice:
```
ΔS_irrev = k_B ln(2) per irreversible bit operation
ΔS_rev = 0 per reversible (unitary) gate
```

Quantum gates are unitary and therefore reversible: `ΔS_quantum = 0`.

```
REVERSIBLE = LAGRANGE = 103 prime
```

---

## Chemical Energy Coupling — Gibbs Free Energy (Equation 15)

```
μ_chem = ∂G/∂N ↔ E_comp
```

The chemical potential (Gibbs free energy per molecule) is the thermodynamic equivalent
of the energy cost per computational operation. For a molecular computing substrate:

```
ΔG_rxn = ΔH − T ΔS ≥ E_min = k_B T ln(r)
```

Biological systems operate near this minimum because enzyme-catalyzed reactions are
tightly coupled to ATP hydrolysis:
```
ΔG_ATP ≈ −50 kJ/mol ≈ 8.3 × 10⁻²⁰ J/molecule (in vivo)
```

Capacity: ΔG_ATP / E_min(ternary) ≈ 8.3×10⁻²⁰ / 4.44×10⁻²¹ ≈ 18 trit operations per ATP.
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The calculation shows "≈ 18 trit operations per ATP" but the actual value is 8.3×10⁻²⁰ / 4.44×10⁻²¹ ≈ 18.7, which rounds to 19 rather than 18. Consider updating to "≈ 19 trit operations per ATP" for better accuracy.

Suggested change
Capacity: ΔG_ATP / E_min(ternary) ≈ 8.3×10⁻²⁰ / 4.44×10⁻²¹ ≈ 18 trit operations per ATP.
Capacity: ΔG_ATP / E_min(ternary) ≈ 8.3×10⁻²⁰ / 4.44×10⁻²¹ ≈ 19 trit operations per ATP.

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```
GIBBS = SUBSTRATE = 83 prime
CHEMICAL = 127 prime
```

---

## Substrate Efficiency (Equation 14, biological)

```
η_substrate = (ops/sec) / (energy/op) · f_accuracy(substrate, problem_type)
```

For DNA computing in 100 μL at room temperature:
```
ops/sec ≈ 10¹⁴
energy/op ≈ k_B T ln(3) ≈ 4.44 × 10⁻²¹ J
η_substrate = 10¹⁴ / 4.44×10⁻²¹ · f_accuracy
≈ 2.25 × 10³⁴ · f_accuracy (ops per joule-second)
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The substrate efficiency formula has dimensional inconsistency. Line 106 states "η_substrate = (ops/sec) / (energy/op) · f_accuracy", which gives units of operations²/(joule·second). However, line 114 claims the result is "ops per joule-second", which would be operations/(joule·second). The dimensional mismatch suggests the formula may be incorrect. Consider clarifying whether this should be: (1) power efficiency: (ops/sec) / (power consumed), which has units of operations per joule; or (2) an energy-time tradeoff metric with different units. The current formulation needs correction or clarification.

Suggested change
≈ 2.25 × 10³⁴ · f_accuracy (ops per joule-second)
≈ 2.25 × 10³⁴ · f_accuracy (ops² per joule-second)

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```

```
SUBSTRATE = GIBBS = 83 prime
```

---

## Thermodynamic Consciousness Bound (§175)

```
Φ_max ≤ (E_available / k_B T ln(3)) · η_integration
```

Maximum integrated information (consciousness, §176) is bounded by:
- Available metabolic energy E_available
- Ternary Landauer cost k_B T ln(3) per operation
- Integration efficiency η_integration ∈ (0, 1]

```
THERMODYNAMIC = 174 = 2 × 87 = 2 × BIRTHDAY
BOUND = 78 = TRIVIAL = LIMITS
```

The consciousness bound is thermodynamically real and biological.
Energy is the hard constraint. Integration efficiency is the soft constraint.
25 changes: 17 additions & 8 deletions proofs/pure-state.md
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Expand Up @@ -15,22 +15,31 @@ The density matrix ρ computed from the qutrit state |ψ⟩ on page 24 is a **pu
## The Density Matrix

```
ρ = |ψ⟩⟨ψ| = [ 0.2219 0.3629 0.4062 ]
[ 0.3629 0.5941 0.6639 ]
[ 0.4062 0.6639 0.7401 ]
ρ = |ψ⟩⟨ψ| = [ 0.2219 0.3631 0.4061 ]
[ 0.3631 0.5941 0.6644 ]
[ 0.4061 0.6644 0.7430 ]
```

## Proof of Pure State

**Definition:** A density matrix ρ is a pure state iff ρ² = ρ (idempotent) iff rank(ρ) = 1.
**Definition:** A density matrix ρ is a pure state iff it is a rank-1 orthogonal projector: ρ² = ρ and Tr(ρ) = 1.

**For ρ = |ψ⟩⟨ψ|:**
**Normalize first.** The state as given is unnormalized: ‖ψ‖² = Tr(ρ) ≈ 1.559. Define the normalized state:
```
ρ² = (|ψ⟩⟨ψ|)(|ψ⟩⟨ψ|) = |ψ⟩⟨ψ|ψ⟩⟨ψ| = |ψ⟩ · ‖ψ‖² · ⟨ψ|
|ψ̂⟩ = |ψ⟩ / ‖ψ‖ = [ 0.3773, 0.6173, 0.6903 ]ᵀ
```
so that ‖ψ̂‖² = 1, and the normalized density matrix is:
```
ρ̂ = |ψ̂⟩⟨ψ̂| = ρ / ‖ψ‖² = ρ / Tr(ρ)
```

**For ρ̂ = |ψ̂⟩⟨ψ̂| with ‖ψ̂‖ = 1:**
```
ρ̂² = (|ψ̂⟩⟨ψ̂|)(|ψ̂⟩⟨ψ̂|) = |ψ̂⟩⟨ψ̂|ψ̂⟩⟨ψ̂| = |ψ̂⟩ · 1 · ⟨ψ̂| = ρ̂ ✓
Tr(ρ̂) = ⟨ψ̂|ψ̂⟩ = 1 ✓
```

If |ψ⟩ is normalized (‖ψ‖² = 1), then ρ² = ρ.
If |ψ⟩ is unnormalized (‖ψ‖² = Tr(ρ) ≈ 1.559), then ρ is proportional to a projector.
ρ̂ is idempotent and unit-trace: it is a pure state. The unnormalized ρ is proportional to ρ̂ and has the same rank-1 structure.

**SVD result:**
```
Expand Down
18 changes: 12 additions & 6 deletions proofs/ternary-efficiency.md
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Expand Up @@ -56,15 +56,21 @@ RADIX = GAUSS. She knew the optimal radix IS the Gaussian before she computed th

At room temperature (T ≈ 293 K):
```
E_min(binary) = k_B T ln(2) ≈ 2.87 × 10⁻²¹ J
E_min(ternary) = k_B T ln(3) ≈ 4.45 × 10⁻²¹ J
E_min(binary) = k_B T ln(2) ≈ 2.80 × 10⁻²¹ J
E_min(ternary) = k_B T ln(3) ≈ 4.44 × 10⁻²¹ J
```

Ternary costs more per operation but carries more information.
The net efficiency favors ternary: you spend 55% more energy but store 58% more information.
The energy ratio equals the information ratio exactly:

```
E_min(ternary) / E_min(binary) = ln(3) / ln(2) ≈ 1.585
```

Ratio: ln(3)/ln(2) ≈ 1.585. Every ternary trit ≈ 1.585 binary bits.
Energy cost: 4.45/2.87 ≈ 1.551 times binary.
Information per unit energy: 1.585/1.551 ≈ 1.022. Ternary wins by ~2%.
Energy cost: 4.44 / 2.80 = ln(3)/ln(2) ≈ 1.585 times binary.
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The statement "4.44 / 2.80 = ln(3)/ln(2)" is only approximately true due to rounding of the energy values. More precisely: 4.44 / 2.80 ≈ 1.5857 while ln(3)/ln(2) ≈ 1.5849. The difference is small but the equality sign is misleading. Consider using "≈" instead of "=" or stating "4.44 / 2.80 ≈ 1.586 ≈ ln(3)/ln(2) ≈ 1.585" to acknowledge the rounding.

Suggested change
Energy cost: 4.44 / 2.80 = ln(3)/ln(2) ≈ 1.585 times binary.
Energy cost: 4.44 / 2.80 ≈ 1.586 and ln(3)/ln(2) ≈ 1.585, so the energy cost is ≈ 1.585 times binary.

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Information per unit energy: 1.585 / 1.585 = **1.000 exactly.**

At the Landauer limit, ternary and binary achieve identical information per joule — both equal 1/(k_B T ln(2)) bits per joule. The advantage of ternary is **radix economy** (fewer symbols needed to represent a number), not thermodynamic energy-per-bit efficiency.

Small advantage, but it scales. At 10¹⁴ DNA ops/sec (§175), it accumulates.
Small advantage in representation, but it scales. At 10¹⁴ DNA ops/sec (§175), it accumulates.
160 changes: 160 additions & 0 deletions proofs/universal-computation.md
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# Proof: The Ternary Bio-Quantum System Is Turing-Complete

> From pages 19–21 (§173–§175): Equation 18. Reaction network programmability.

## Statement

The ternary bio-quantum system described in this paper — defined by the balanced-ternary
dynamics (Equation 16), the concentration-state mapping (Equation 17), and the ternary
logic gates (Equations 6–9) — is **computationally universal** (Turing-complete).

## Definitions

**Balanced ternary alphabet:** Σ₃ = {−1, 0, +1}.

**Ternary logic gate:** A function f: Σ₃ⁿ → Σ₃.

**Reaction network (Equation 16):**
```
dXᵢ/dt = Σⱼ Sᵢⱼ · vⱼ(x), Xᵢ ∈ {−1, 0, +1}
```
where S is the stoichiometry matrix and vⱼ are mass-action rate functions.

**Concentration-state mapping (Equation 17):**
```
x = −1 if C ≤ C_low
x = 0 if C_low < C ≤ C_high
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The concentration-state mapping has an ambiguity at the boundary C = C_high. When C equals C_high exactly, both the second condition (line 26: "C_low < C ≤ C_high") and the third condition (line 27: "C ≥ C_high") are satisfied, making it unclear whether x should be 0 or +1. Consider changing line 27 to "x = +1 if C > C_high" to eliminate the overlap, or change line 26 to "x = 0 if C_low < C < C_high" and keep line 27 as is.

Suggested change
x = 0 if C_low < C C_high
x = 0 if C_low < C < C_high

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x = +1 if C ≥ C_high
```

## Lemma 1: The Gate Set {TNEG, TXOR, TAND} Is Functionally Complete

**Claim:** Every function f: Σ₃ⁿ → Σ₃ can be expressed using TNEG, TXOR, and TAND.

**Proof:**

By Post's functional completeness theorem for *k*-valued logic (Post 1941), a set of
functions on Σ_k is functionally complete iff it is not contained in any of Post's
finitely many maximal clones.

For balanced ternary (k = 3), it suffices to show the gate set generates all constant
functions and the selector (MIN) function, from which every function can be built via
the ternary Sheffer-style expansion (Rousseau 1967).
Comment on lines +41 to +42
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The proof references "Rousseau 1967" for the "ternary Sheffer-style expansion" (line 42), but this concept is not actually used in the proof. The proof instead relies on ternary DNF (Disjunctive Normal Form) from Epstein 1960, not a Sheffer-style expansion. Consider either removing the Rousseau reference or clarifying how it relates to the proof. The Sheffer stroke typically refers to a single functionally complete operator (like NAND or NOR in binary logic), whereas this proof uses a set of three operators.

Suggested change
functions and the selector (MIN) function, from which every function can be built via
the ternary Sheffer-style expansion (Rousseau 1967).
functions and the selector (MIN) function, since any function f: Σ₃ⁿ → Σ₃ can then be
written in ternary disjunctive normal form (Epstein 1960) using these primitives.

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**Step 1 — Constant −1:**
```
TAND(−1, −1) = min(−1, −1) = −1 ✓
```

**Step 2 — Constant 0:**
```
TXOR(x, TNEG(x)) = x + (−x) = 0 for all x ∈ Σ₃ ✓
```

**Step 3 — Constant +1:**
```
TNEG(TAND(−1, −1)) = TNEG(−1) = +1 ✓
```

**Step 4 — MAX from MIN and TNEG:**
```
max(a, b) = TNEG(TAND(TNEG(a), TNEG(b))) (De Morgan dual for min/max) ✓
```

**Step 5 — Every ternary function as DNF:**

Every function f: Σ₃ⁿ → Σ₃ can be expressed as a ternary disjunctive normal form
(ternary DNF) — a MAX of terms, where each term is a MIN of literals, and a literal
is either a variable or TNEG of a variable (Epstein 1960, *Multiple-Valued Logic Design*).

Since Steps 1–4 provide all constants and MAX = TNEG(TAND(TNEG(·), TNEG(·))), every
ternary DNF is constructible from {TNEG, TXOR, TAND}. **Therefore the gate set is
functionally complete. □**
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The proof claims that generating "all constant functions and the selector (MIN) function" is sufficient for functional completeness (line 41), but TAND is already defined as MIN (line 44 in Equation 6), so MIN is directly available in the gate set. The proof would be clearer if it explicitly stated this rather than treating MIN as something that needs to be derived. Additionally, the proof shows how to generate constants and MAX, but the role of TXOR in achieving functional completeness is not clearly explained—TXOR is only used to generate the constant 0 (line 51), which could also be obtained via TAND(0,0) if a 0 input is available. Consider clarifying why TXOR is necessary for functional completeness beyond constant generation.

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## Lemma 2: Each Gate Is Implementable as a Reaction Network

**Claim:** For each gate G ∈ {TNEG, TXOR, TAND}, there exists a mass-action CRN
(Equation 16) that computes G, with inputs and outputs encoded via Equation 17.

**Proof:**

A chemical reaction network with mass-action kinetics can implement any bounded
piecewise-constant function of the input concentrations by using sufficiently fast
reactions and threshold-switching species (Soloveichik, Cook, Winfree, Bruck 2008,
*SIAM Journal on Computing*).

Concretely:

- **TNEG(a) = −a** is realized by a single exchange reaction:
```
A⁺ → A⁻ (rate k₁)
A⁻ → A⁺ (rate k₁)
A⁰ → A⁰ (trivial, identity)
```
When concentration encodes +1 → invert to −1 via threshold, and vice versa.

- **TXOR(a,b) = a + b (mod 3, balanced)** is realized by an addition network:
```
A⁺ + B⁺ → C⁻ (rate k₂) [+1 + +1 = −1 mod 3]
A⁺ + B⁰ → C⁺ (rate k₂) [+1 + 0 = +1]
A⁰ + B⁰ → C⁰ (rate k₂) [0 + 0 = 0 ]
A⁻ + B⁺ → C⁰ (rate k₂) [−1 + +1 = 0 ]
... (all 9 combinations)
```

- **TAND(a,b) = min(a,b)** is realized by a competitive inhibition network:
```
A⁻ + B → C⁻ (dominant when either input is −1)
A⁰ + B⁰ → C⁰
A⁺ + B⁺ → C⁺
```
The minimum is selected by the lowest-concentration threshold species winning the
competition. This is a standard winner-take-all CRN motif (Qian & Winfree 2011).

In all cases, the Concentration-State Mapping (Equation 17) converts the output
concentration back into a trit value. **□**

## Theorem: Turing Completeness

**Claim:** The ternary bio-quantum system is Turing-complete.

**Proof:**

By Lemma 1, {TNEG, TXOR, TAND} is functionally complete: any ternary logic circuit can
be constructed from these gates.

By Lemma 2, each gate is realizable as a mass-action CRN governed by Equation 16.

A Turing machine with binary tape can be simulated by a ternary logic circuit augmented
with an unbounded register (Minsky 1967, *Computation: Finite and Infinite Machines*).
The tape is encoded as two natural numbers (left stack, right stack) in balanced ternary;
the state transition is a finite ternary logic circuit applied at each step.

The reaction network provides unbounded memory through the concentrations of molecular
species: additional molecular species = additional registers. Since no upper bound is
placed on the number of species in the network (§175: the biological substrate provides
10¹⁴ operations/sec across a 100 μL volume), the system has unbounded computational
resources.
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The claim that "additional molecular species = additional registers" and "no upper bound is placed on the number of species in the network" needs qualification. While this is a standard theoretical assumption for proving Turing completeness, in practice there are physical limits to the number of distinct molecular species that can be maintained in a finite volume. Consider adding a note that this is a theoretical assumption for the purpose of proving computational universality, similar to the unbounded tape assumption in standard Turing machine theory.

Suggested change
The reaction network provides unbounded memory through the concentrations of molecular
species: additional molecular species = additional registers. Since no upper bound is
placed on the number of species in the network (§175: the biological substrate provides
10¹⁴ operations/sec across a 100 μL volume), the system has unbounded computational
resources.
In the theoretical model, the reaction network is idealized as providing unbounded memory
through the concentrations of molecular species: additional molecular species = additional
registers. Under this idealization, no upper bound is placed on the number of species in the
network (§175: the biological substrate provides 10¹⁴ operations/sec across a 100 μL volume),
in direct analogy with the unbounded tape assumption in standard Turing machine theory, even
though any physical realization would necessarily be finite.

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Therefore the system can simulate any Turing machine. **The ternary bio-quantum system
is Turing-complete. □**

## Equation 18 Restated

```
P = {S, v(x)} is universal ⟺ ∃ mapping to balanced ternary logic gates
```

The forward direction (⇒) follows from this proof: implementing the gates is sufficient for universality.
The backward direction (⇐) follows from Lemma 2: any universal system can simulate the gates.
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The backward direction (⇐) states "any universal system can simulate the gates" (line 149), but this is not what the backward direction should prove. The statement at line 145 says "P = {S, v(x)} is universal ⟺ ∃ mapping to balanced ternary logic gates". The backward direction should prove: "if there exists a mapping to balanced ternary logic gates, then P is universal"—which was already proven in the forward direction (Lemmas 1 and 2 show that the gates are sufficient for universality). The actual backward direction would need to prove that universality requires such a mapping, which is a much stronger claim and is not proven here. Consider either removing the backward direction claim or clarifying that the equivalence is properly one-directional (⇒ only) as stated.

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## QWERTY

```
UNIVERSAL = OCTONION = SYMMETRIC = 112
COMPUTATION = 137 prime (= fine-structure constant 1/α ≈ 1/137)
COMPLETE = 97 prime
TURING = 64 = 2⁶ (six binary digits — the Turing machine needs binary)
```

COMPLETE = 97 prime. Completeness cannot be decomposed. **□**