q=3 Next-Digit Normal Form — Free Computation Foundation

Theorem

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Status: proved universal theorem

Date: 2026-08-15

Depends on: REDUCED-PARAMETER-DOMAIN.md, Q3-FIBER-INJECTIVITY.md

Claim boundary: normalizes every q=3 pullback on a fixed candidate into one common next 3-adic digit coordinate. It does not prove that three primitive base rows cannot occupy all three digits, universal DSC-P, López-all-primes, or Erdős-Straus.

1. Setup

Fix a candidate progression

x(s)=r+Ls.

Let j be an earlier layer with

m_j=4j-1

and

q_j=\frac{m_j}{\gcd(L,m_j)}=3.

Write

\boxed{n_j=m_j/3=\gcd(L,m_j).}

Let

b=v_3(n_j).

Because n_j|L but 3n_j does not divide L,

\boxed{v_3(L)=b.}

Thus every q=3 layer active on the same candidate has the same value of b.

Write

n_j=3^b\nu_j, \qquad L=3^b\Lambda,

with

3\nmid\nu_j\Lambda.

Since n_j|L, we also have \nu_j|\Lambda.

2. Theorem — layer modulus cancels

Suppose a trap u in T_j produces a nonempty pullback class

a\in R_j\subseteq\mathbb Z/3\mathbb Z.

Then

\boxed{ a\equiv \frac{u-r}{3^b}\,\Lambda^{-1} \pmod3. }

Equivalently,

\boxed{ a\equiv \left(\frac{u-r}{3^b}\right) \left(\frac{L}{3^b}\right)^{-1} \pmod3. }

Proof

Since q_j=3, the hit relation is

r+La\equiv u\pmod{3n_j}.

Reducing modulo n_j gives

u\equiv r\pmod{n_j},

D:=\frac{u-r}{n_j}

is an integer, understood modulo 3.

Divide the hit relation by n_j:

\frac{L}{n_j}a\equiv D\pmod3.

Because q_j=3, L/n_j is a unit modulo 3, hence

a\equiv D\left(\frac{L}{n_j}\right)^{-1}\pmod3.

Now

D =\frac{u-r}{3^b\nu_j} \equiv \frac{u-r}{3^b}\,\nu_j^{-1} \pmod3,

while

\left(\frac{L}{n_j}\right)^{-1} = \left(\frac{\Lambda}{\nu_j}\right)^{-1} \equiv \nu_j\Lambda^{-1} \pmod3.

The layer-specific factor \nu_j cancels:

a\equiv \frac{u-r}{3^b}\,\Lambda^{-1} \pmod3.

QED.

3. Interpretation — every row paints one next 3-adic digit

The condition

u\equiv r\pmod{n_j}

already implies

u\equiv r\pmod{3^b}.

So a q=3 trap witness lies above the fixed 3^b-adic prefix of the candidate and chooses one of the three lifts modulo 3^{b+1}.

The theorem says that the parameter class a mod 3 is exactly that next digit, followed by one fixed multiplication by the common unit

\Lambda^{-1}=(L/3^b)^{-1}\pmod3.

Therefore all q=3 layers on the candidate use the same coordinate system. No layer-specific rescaling remains.

4. Cover criterion in digit form

For a q=3 trap witness u, define its normalized next digit

\boxed{ \delta(u):= \frac{u-r}{3^b}\pmod3. }

Multiplication by \Lambda^{-1} is a permutation of F_3, so

\bigcup_jR_j=\mathbb Z/3\mathbb Z

if and only if the realized trap witnesses occupy all three normalized next digits:

\boxed{ \{\delta(u):u\text{ is a realized q=3 trap witness}\} =\mathbb F_3. }

Combined with Q3-FIBER-INJECTIVITY.md, each individual q=3 layer contributes at most one such digit.

5. Directly novel form

Combine the q=3 hierarchy:

strong descendants are absent when nonempty on a directly novel candidate;
weak descendants are redundant;
every used trap must be pointwise primitive;
every surviving layer contributes at most one class;
every surviving class is one common next 3-adic digit.

Thus a genuine directly novel q=3 obstruction is precisely:

\boxed{ \text{at least three pointwise-primitive base trap witnesses} \text{ occupying all three lifts above one }3^b\text{-prefix}. }

This is the exact residual local theorem target.

6. Special case b=1

When v_3(L)=1, every hard candidate satisfies r=1 mod 3, and the three possible trap residues modulo 9 are

1,4,7.

A corrected q=3 cover is therefore equivalent to primitive realized traps occupying all three residues

\boxed{1,4,7\pmod9}

(up to the common affine permutation determined by r and L/3).

This gives a concrete small-modulus target for the base-triple analysis.