q=3 Weak-Ancestor Redundancy — Free Computation Foundation

Theorem

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Status: proved universal theorem

Date: 2026-08-15

Depends on: Q3-ABSORPTION.md

Claim boundary: reduces the corrected q=3 shared-factor obstruction to base layers. Does not prove that base layers cannot cover the corrected parameter domain, universal DSC-P, López-all-primes, or Erdős-Straus.

Setup

Let

m_j=4j-1, \qquad q_j=\frac{m_j}{\gcd(L,m_j)}.

Call a layer j weak when there is an earlier layer i<j such that

\boxed{m_i\mid m_j}

and

\boxed{T_j\bmod m_i\subseteq T_i,}

but

\boxed{m_i\nmid m_j/3.}

Thus j has a trap-reducing ancestor, but not a strong-absorption ancestor in the sense of Q3-ABSORPTION.md.

Theorem (weak q=3 redundancy)

Assume

\boxed{q_j=3.}

If i is a weak trap-reducing ancestor of j, then

\boxed{q_i=3}

and, for the same candidate progression,

\boxed{R_j\subseteq R_i\subseteq\mathbb Z/3\mathbb Z.}

Therefore a weak q=3 layer contributes no forbidden parameter residue not already contributed by an earlier layer.

Proof

Put

g=\gcd(L,m_j).

Since q_j=3,

\boxed{m_j=3g}

and g|L.

Because m_i|m_j, every common divisor of L and m_i is also a common divisor of L and m_j; hence

\gcd(L,m_i)=\gcd(g,m_i).

Now m_i|3g. Therefore

\frac{m_i}{\gcd(g,m_i)}\mid3.

The weak hypothesis says m_i does not divide g=m_j/3, so this quotient is greater than 1. Since 3 is prime,

\boxed{ \frac{m_i}{\gcd(L,m_i)} = \frac{m_i}{\gcd(g,m_i)} =3. }

Thus q_i=3.

Now take any residue class

a\in R_j.

By definition there is a parameter s with

s\equiv a\pmod3

such that

x(s)=r+Ls\pmod{m_j}\in T_j.

Reducing modulo m_i and using

T_j\bmod m_i\subseteq T_i

gives

x(s)\pmod{m_i}\in T_i.

Since q_i=3, this means exactly

s\bmod3\in R_i.

Hence a in R_i, proving

\boxed{R_j\subseteq R_i.}

QED.

Corollary 1 — weak layers can be deleted from q=3 covers

For any candidate progression, repeatedly delete every weak q=3 layer and retain one of its earlier weak ancestors. The union of forbidden classes modulo 3 does not shrink:

\boxed{ \bigcup_{j\in\mathcal W}R_j \subseteq \bigcup_{i\in\mathcal A}R_i, }

where A contains earlier ancestors reached by the reductions.

Thus weak descendants are redundant for every q=3 covering question.

Corollary 2 — directly novel corrected q=3 obstruction = base-only obstruction

Combine this theorem with Strong q=3 absorption:

strong layer: if R_j != empty, the candidate is directly shadowed and therefore cannot occur on a directly novel candidate;
weak layer: R_j is contained in an earlier q=3 ancestor pullback and adds no new forbidden class;
base layer: no trap-reducing ancestor is available from this hierarchy.

Therefore on directly novel candidates, every genuinely new forbidden residue in the q=3 coordinate can be represented by a base q=3 layer.

In the exact Dirichlet parameter domain, 3|840|L, so the local domain is all of

\mathbb Z/3\mathbb Z.

A genuine q=3 local obstruction must therefore reduce to base layers whose pullbacks cover all three classes

\boxed{\{0,1,2\}.}

The residual theorem target is no longer strong+weak+base, and no longer a complementary pair. It is:

\boxed{ \text{Can base q=3 layers cover }\mathbb Z/3\mathbb Z \text{ on a directly novel admissible candidate?} }

Consequence for the j<=1500 ancestry census

Q3-ABSORPTION.md records:

strong absorb: 153
weak only:      114
base:           233

The first two populations are now structurally removed from the novel q=3 obstruction:

153 strong layers are novel-impossible when active;
114 weak layers are residue-redundant;
only the 233 base layers can contribute genuinely new q=3 forbidden classes in that finite census.

The counts are finite census data; the strong/weak reductions themselves are universal.