> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# Reverse Goldbach Structural Closure Program 02
## Goldbach Failure Obstruction Cover: From Prime-Complement Failure to Least-Prime-Factor Partition
### Reverse Goldbach Structural Closure Program 02
### Goldbach Failure Obstruction Cover: From Prime-Complement Failure to Least-Prime-Factor Partition

**Series:** Reverse Goldbach Structural Closure Program  
**English name:** Reverse Goldbach Structural Closure Program  
**Abbreviation:** RG-SCP  
**Paper:** 02 / planned 08–10  
**Authors:** Neo.K; Aletheia (GPT-5.6 Sol)  
**Date:** 2026-09-28  
**Version:** v0.1  
**Document type:** Research program / structural-theorem draft / preliminary document for subsequent M6, Wheel, and Fixed-Point layers  
**Status:** Not a proof of the Goldbach conjecture; this paper establishes an exact obstruction-cover equivalence for Goldbach failure

---

## Abstract

The first paper of the Reverse Goldbach Structural Closure Program changed the object of research on binary Goldbach from 'directly proving that every even number is a sum of two primes' to 'compressing the structurally admissible space of all possible Goldbach failure states'.

This paper completes the first genuinely structural step:

> If an even number $N$ is a Goldbach failure state, then for every prime $p$ available as the first summand, its complement $N-p$ must be composite; thus every $p$ must be blocked by at least one proper prime factor of $N-p$.

This leads to an exact obstruction-cover representation.

Let:

$$
\mathbb P_N
=
\left\{
p\in\mathbb P:
2\le p\le N-2
\right\}.
$$

If $N$ is a Goldbach failure, then for every $p\in\mathbb P_N$:

$$
N-p
$$

is composite, so there exists a prime $r$ such that:

$$
r\mid(N-p),
$$

and:

$$
r\le\sqrt{N-p}\le\sqrt{N-2}.
$$

Equivalently:

$$
p\equiv N\pmod r.
$$

Thus one can define the obstruction class:

$$
\mathcal O_r(N)
=
\left\{
p\in\mathbb P_N:
p\equiv N\pmod r
\right\},
$$

and obtain the necessary cover:

$$
\boxed{
\mathbb P_N
\subseteq
\bigcup_{\substack{
r\in\mathbb P\\
r\le\sqrt{N-2}
}}
\mathcal O_r(N).
}
$$

This paper further proves a more precise equivalent form:

$$
\boxed{
F_G(N)=1
}
$$

if and only if there exists a total proper obstruction map:

$$
\Phi_N:
\mathbb P_N
\to
\mathbb P
$$

such that every $p\in\mathbb P_N$ satisfies:

$$
\Phi_N(p)\mid(N-p),
$$

and:

$$
2\le
\Phi_N(p)
\le
\sqrt{N-p}.
$$

If one takes:

$$
\Phi_N^{\min}(p)
=
\operatorname{lpf}(N-p),
$$

where $\operatorname{lpf}$ denotes the least prime factor, one obtains a canonical obstruction map and partitions the prime domain into disjoint least-factor obstruction cells.

This paper also establishes:

- full obstruction cover；
- reduced half-domain cover；
- least-prime-factor partition；
- obstruction depth；
- wheel-truncated residual；
- divisor-of-$N$ obstruction suppression；
- $M_6$ easy/hard branch；
- source-class audit；
- a serialization schema.

The core research transformation is:

$$
\boxed{
\text{Goldbach failure}
\Longleftrightarrow
\text{existence of a complete proper obstruction assignment}.
}
$$

This is not a Goldbach proof, but it compresses a counterexample from 'some mysterious even number' into an auditable structure required to cover the entire prime-complement domain.

---

# 1. From a Failure State to an Obstruction Obligation

Define the standard binary Goldbach representation count:

$$
R_G(N)
=
\#\left\{
p\in\mathbb P:
2\le p\le N-2,
\quad
N-p\in\mathbb P
\right\}.
$$

The Goldbach failure predicate is:

$$
F_G(N)
=
1
\Longleftrightarrow
R_G(N)=0.
$$

If:

$$
F_G(N)=1,
$$

then:

$$
\forall p\in\mathbb P_N,
\quad
N-p\notin\mathbb P.
$$

Because:

$$
2\le N-p\le N-2,
$$

so $N-p$ is neither $1$ nor $0$.

Therefore:

$$
\boxed{
\forall p\in\mathbb P_N,
\quad
N-p
\text{ must be composite}.
}
$$

This is the starting point for every obstruction structure in this paper.

---

# 2. The Proper-Prime-Factor Obligation

For any composite $m\ge4$, there exists a prime $r$ such that:

$$
r\mid m
$$

and:

$$
r\le\sqrt m.
$$

Therefore, if:

$$
F_G(N)=1,
$$

then for any:

$$
p\in\mathbb P_N,
$$

there exists a prime $r$ such that:

$$
r\mid(N-p)
$$

and:

$$
r\le\sqrt{N-p}.
$$

So:

$$
r\le\sqrt{N-2}.
$$

This means that every Goldbach failure produces a finite obstruction-prime universe:

$$
\mathcal R_N
=
\left\{
r\in\mathbb P:
r\le\sqrt{N-2}
\right\}.
$$

---

# 3. Congruence obstruction

From:

$$
r\mid(N-p)
$$

we immediately obtain:

$$
N-p\equiv0\pmod r,
$$

Thus:

$$
\boxed{
p\equiv N\pmod r.
}
$$

Thus every complement-factor obstruction can be projected to a residue obstruction.

Define:

$$
\mathcal O_r(N)
=
\left\{
p\in\mathbb P_N:
r\mid(N-p)
\right\}.
$$

Equivalently:

$$
\mathcal O_r(N)
=
\left\{
p\in\mathbb P_N:
p\equiv N\pmod r
\right\}.
$$

An obstruction is therefore not an arbitrary label attached to a prime.

It must share the same residue relation with $N$.

---

# 4. Goldbach Failure Obstruction Cover

If:

$$
F_G(N)=1,
$$

then every:

$$
p\in\mathbb P_N
$$

belongs to at least one:

$$
\mathcal O_r(N),
$$

where:

$$
r\in\mathcal R_N.
$$

Therefore:

$$
\boxed{
\mathbb P_N
\subseteq
\bigcup_{r\in\mathcal R_N}
\mathcal O_r(N).
}
$$

Call this:

# Goldbach Failure Obstruction Cover

Abbreviated:

$$
\boxed{
\operatorname{GFOC}(N).
}
$$

This cover is a necessary condition for Goldbach failure.

---

# 5. Why a Cover Alone Is Not Enough

A complement:

$$
N-p
$$

may have several distinct prime factors.

For example:

$$
N-p
=
r_1r_2r_3.
$$

Then the same $p$ simultaneously belongs to:

$$
\mathcal O_{r_1}(N),
\quad
\mathcal O_{r_2}(N),
\quad
\mathcal O_{r_3}(N).
$$

Thus raw GFOC is highly overlapping.

If subsequent research is to examine:

- cover cost；
- obstruction depth；
- branch load；
- serialization；
- source provenance；

a canonical assignment is needed.

---

# 6. Least Prime Factor

For composite $m$, define:

$$
\operatorname{lpf}(m)
=
\min
\left\{
r\in\mathbb P:
r\mid m
\right\}.
$$

If $m$ is composite, then:

$$
\operatorname{lpf}(m)
\le
\sqrt m.
$$

Therefore, if:

$$
F_G(N)=1,
$$

one can, for every:

$$
p\in\mathbb P_N
$$

Define:

$$
\boxed{
\Phi_N^{\min}(p)
=
\operatorname{lpf}(N-p).
}
$$

This is called the:

# Canonical Least-Factor Obstruction Map

---

# 7. Total Proper Obstruction Map

More generally, define a map:

$$
\Phi_N:
\mathbb P_N
\to
\mathbb P.
$$

If every $p\in\mathbb P_N$ satisfies:

$$
\Phi_N(p)\mid(N-p)
$$

and:

$$
\Phi_N(p)
\le
\sqrt{N-p},
$$

then:

$$
\Phi_N
$$

is called a **total proper obstruction map**.

The importance of 'proper' is that:

If:

$$
N-p
$$

is itself prime, its only prime factor is:

$$
N-p,
$$

and it does not satisfy:

$$
N-p
\le
\sqrt{N-p}
$$

except at nonexistent trivial small values.

Thus the proper-factor bound automatically excludes 'using the complement itself as a spurious obstruction'.

---

# 8. Obstruction Map Characterization Theorem

## Theorem RG-SCP-02.1

For even:

$$
N\ge4,
$$

the following two statements are equivalent:

### A

$$
F_G(N)=1.
$$

### B

There exists a total proper obstruction map:

$$
\Phi_N:
\mathbb P_N
\to
\mathbb P
$$

such that for every:

$$
p\in\mathbb P_N
$$

we have:

$$
\Phi_N(p)\mid(N-p)
$$

and:

$$
\Phi_N(p)
\le
\sqrt{N-p}.
$$

---

## Proof

### A Implies B

Suppose:

$$
F_G(N)=1.
$$

Then for every:

$$
p\in\mathbb P_N,
$$

the complement:

$$
N-p
$$

is composite.

Thus:

$$
\operatorname{lpf}(N-p)
$$

exists, with:

$$
\operatorname{lpf}(N-p)
\le
\sqrt{N-p}.
$$

Let:

$$
\Phi_N(p)
=
\operatorname{lpf}(N-p).
$$

yielding a total proper obstruction map.

### B Implies A

Conversely, suppose such a:

$$
\Phi_N.
$$

Then for every:

$$
p\in\mathbb P_N,
$$

there is a prime:

$$
\Phi_N(p)
$$

such that:

$$
\Phi_N(p)\mid(N-p)
$$

and:

$$
1<
\Phi_N(p)
<
N-p.
$$

Therefore:

$$
N-p
$$

must be composite.

Thus there is no:

$$
p\in\mathbb P_N
$$

such that:

$$
N-p\in\mathbb P.
$$

Hence:

$$
R_G(N)=0.
$$

Therefore:

$$
F_G(N)=1.
$$

This completes the proof.

---

# 9. Significance of This Equivalent Form

The Goldbach conjecture can be rewritten as:

$$
\boxed{
\forall N\ge4,
\quad
N\equiv0\pmod2
\Longrightarrow
\text{there is no total proper obstruction map } \Phi_N.
}
$$

The Goldbach problem thus acquires an entirely new reverse entry point:

> Rather than seeking a prime pair directly, investigate whether a proper-factor obstruction assignment can be completed for all prime inputs.

That is:

$$
\text{Prime Pair Existence}
$$

is transformed into:

$$
\text{Complete Obstruction Assignment Impossibility}.
$$

---

# 10. Canonical Obstruction Partition

Using:

$$
\Phi_N^{\min}(p)
=
\operatorname{lpf}(N-p),
$$

For each:

$$
r\in\mathcal R_N
$$

Define:

$$
\mathcal O_r^{\min}(N)
=
\left\{
p\in\mathbb P_N:
\Phi_N^{\min}(p)=r
\right\}.
$$

By uniqueness of the least prime factor:

$$
\mathcal O_r^{\min}(N)
\cap
\mathcal O_s^{\min}(N)
=
\varnothing
$$

For:

$$
r\neq s.
$$

And if:

$$
F_G(N)=1,
$$

then:

$$
\boxed{
\mathbb P_N
=
\bigsqcup_{r\in\mathcal R_N}
\mathcal O_r^{\min}(N).
}
$$

the raw cover is thus upgraded to a canonical partition.

---

# 11. Obstruction Coloring

One can regard:

$$
\Phi_N^{\min}
$$

as a coloring of the prime domain:

$$
p
\mapsto
r.
$$

where the color $r$ indicates:

> Which least prime factor first certifies that $N-p$ is composite.

Thus a hypothetical Goldbach counterexample must produce a complete coloring:

$$
\boxed{
\mathbb P_N
\to
\mathcal R_N.
}
$$

Moreover, this coloring is not arbitrary.

Every color $r$ must satisfy:

$$
p\equiv N\pmod r.
$$

So this is a:

$$
\boxed{
\text{congruence-constrained prime coloring}.
}
$$

---

# 12. Reduced Half-Domain

The full:

$$
\mathbb P_N
$$

includes both sides of a complementary pair.

If:

$$
p+q=N,
$$

then at least one satisfies:

$$
p\le\frac N2.
$$

Thus one can define the reduced prime domain:

$$
\mathbb P_N^{-}
=
\left\{
p\in\mathbb P:
2\le p\le\frac N2
\right\}.
$$

then:

$$
F_G(N)=1
$$

is equivalent to:

$$
\forall p\in\mathbb P_N^{-},
\quad
N-p
\notin\mathbb P.
$$

Therefore all obstruction theorems can be established only on:

$$
\mathbb P_N^{-}
$$

the domain displayed above.

This roughly halves the proof obligation without changing the proposition's meaning.

---

# 13. Reduced Obstruction Characterization

Define:

$$
\Phi_N^{-}:
\mathbb P_N^{-}
\to
\mathbb P.
$$

If:

$$
\Phi_N^{-}(p)\mid(N-p)
$$

and:

$$
\Phi_N^{-}(p)
\le
\sqrt{N-p},
$$

for all:

$$
p\in\mathbb P_N^{-}
$$

holds for all such inputs, then:

$$
F_G(N)=1.
$$

and conversely.

Thus:

$$
\boxed{
F_G(N)=1
\Longleftrightarrow
\exists
\text{ total proper reduced obstruction map}.
}
$$

Subsequent experiments and formalization can prioritize the reduced domain.

---

# 14. Obstruction Load

For the canonical partition, define the load of each obstruction prime:

$$
L_r(N)
=
\left|
\mathcal O_r^{\min}(N)
\right|.
$$

If:

$$
F_G(N)=1,
$$

then:

$$
\sum_{r\in\mathcal R_N}
L_r(N)
=
|\mathbb P_N|.
$$

In the reduced domain:

$$
\sum_{r\in\mathcal R_N}
L_r^{-}(N)
=
|\mathbb P_N^{-}|.
$$

This allows a counterexample to be represented by an obstruction-load spectrum:

$$
\boxed{
\mathcal L_N
=
\left(
L_2(N),
L_3(N),
L_5(N),
L_7(N),
\ldots
\right).
}
$$

---

# 15. Obstruction Depth

Define the canonical obstruction depth:

$$
D_G(N)
=
\max_{p\in\mathbb P_N^{-}}
\Phi_N^{\min}(p).
$$

If:

$$
F_G(N)=1,
$$

then:

$$
D_G(N)
\le
\sqrt{N-2}.
$$

$D_G(N)$ means:

> How large a least prime factor must be reached at the deepest point to prove that all reduced prime complements are composite.

If:

$$
D_G(N)
$$

is close to:

$$
\sqrt N,
$$

the obstruction cover nearly requires the full trial-factor depth.

If:

$$
D_G(N)
$$

is much smaller than:

$$
\sqrt N,
$$

then the failure state has achieved complete coverage at a very shallow factor scale.

This is important for subsequently distinguishing 'genuine structural compression' from 'exhaustive blocking'.

---

# 16. Primorial Truncation

Let:

$$
P(y)
=
\prod_{\substack{
r\le y\\
r\in\mathbb P
}}
r.
$$

Define the $y$-level obstruction-covered prime set:

$$
\mathbb P_N^{\mathrm{cov}}(y)
=
\left\{
p\in\mathbb P_N^{-}:
\gcd(N-p,P(y))>1
\right\}.
$$

and the residual survivor set:

$$
\mathbb P_N^{\mathrm{res}}(y)
=
\left\{
p\in\mathbb P_N^{-}:
\gcd(N-p,P(y))=1
\right\}.
$$

Therefore:

$$
\mathbb P_N^{-}
=
\mathbb P_N^{\mathrm{cov}}(y)
\bigsqcup
\mathbb P_N^{\mathrm{res}}(y).
$$

---

# 17. Truncated Cover Ratio

Define:

$$
\Theta_N(y)
=
\frac{
|\mathbb P_N^{\mathrm{cov}}(y)|
}{
|\mathbb P_N^{-}|
}.
$$

Then:

$$
0\le\Theta_N(y)\le1.
$$

And:

$$
y_1\le y_2
\Longrightarrow
\Theta_N(y_1)
\le
\Theta_N(y_2).
$$

If:

$$
F_G(N)=1,
$$

then when:

$$
y\ge D_G(N)
$$

as displayed above:

$$
\Theta_N(y)=1.
$$

Therefore:

$$
D_G(N)
=
\min
\left\{
y:
\Theta_N(y)=1
\right\}
$$

If $y$ is restricted to prime levels, a corresponding discretization can be used.

---

# 18. Why $\sqrt N$ Closure Is Not New Proof Compression

For any composite:

$$
m\le N,
$$

checking all:

$$
r\le\sqrt m,
$$

finds a prime factor.

Thus, if Goldbach failure has already been assumed, obtaining a complete obstruction cover when:

$$
y\sim\sqrt N
$$

as displayed above mainly amounts to:

$$
\text{every complement is composite}
$$

rephrasing this as:

$$
\text{every complement has a factor below its square root}.
$$

This is not new proof compression.

What is genuinely worth investigating is:

> Is there a uniform structural mechanism, far below the full factor-search depth, that can force a hypothetical failure to complete the cover?

That is, seek:

$$
y(N)\ll\sqrt N
$$

or a higher-order closure mechanism not consisting merely of a factor cutoff.

---

# 19. A Fixed-Period Source Warning

If $y$ is fixed, then:

$$
P(y)
$$

is a fixed modulus.

For:

$$
\gcd(N-p,P(y))
$$

the observations still belong to a finite fixed-period source.

Thus, following the warning of the Prime-Surface Insufficiency line:

$$
\boxed{
\text{a wheel cover with fixed }y
\text{ should not be expected to close global Goldbach on its own}.
}
$$

If one merely keeps taking:

$$
y
\to
\text{a larger fixed value},
$$

that may only be:

$$
\text{deeper exhaustive refinement}.
$$

RG-SCP aims not to make the wheel infinitely deep, but to determine when one must leave that source class.

---

# 20. Divisor-of-$N$ Obstruction Suppression

Here the first nontrivial yet simple structural restriction appears.

## Theorem RG-SCP-02.2

Let:

$$
r
$$

be prime, with:

$$
r\mid N.
$$

If:

$$
p\in\mathbb P
$$

and:

$$
r\mid(N-p),
$$

then:

$$
p=r.
$$

---

## Proof

From:

$$
r\mid N
$$

and:

$$
r\mid(N-p),
$$

we obtain:

$$
r\mid p.
$$

Since $p$ is prime:

$$
p=r.
$$

This completes the proof.

---

# 21. Self-Suppression Principle

Therefore, if:

$$
r\mid N,
$$

then the obstruction class:

$$
\mathcal O_r(N)
$$

can cover at most:

$$
p=r.
$$

That is:

$$
\boxed{
r\mid N
\Longrightarrow
\mathcal O_r(N)
\subseteq
\{r\}.
}
$$

This means:

> If $N$ itself contains a small prime factor, that factor loses the ability to cover other prime inputs.

We call this:

# Obstruction Self-Suppression

---

# 22. The Intuitive Meaning of Self-Suppression

If a hypothetical counterexample $N$ contains many small prime factors:

$$
2,3,5,\ldots,
$$

then, for most:

$$
p\in\mathbb P_N
$$

these small prime factors cannot serve as:

$$
N-p
$$

an obstruction factor of the complement displayed above.

Therefore:

$$
N
$$

the more small-prime smoothness it has, the more it may need to push the obstruction burden toward larger prime factors.

This is not a Goldbach proof.

But it supplies a new measurable pressure:

$$
\boxed{
\text{small factors of }N
\quad\text{vs.}\quad
\text{available small obstruction channels}.
}
$$

---

# 23. Prime Divisor Suppression Set

Define:

$$
\mathcal D(N)
=
\left\{
r\in\mathbb P:
r\mid N
\right\}.
$$

Then, for:

$$
p\in\mathbb P_N
\setminus
\mathcal D(N),
$$

any canonical obstruction:

$$
\Phi_N^{\min}(p)
$$

cannot belong to:

$$
\mathcal D(N).
$$

Therefore:

$$
\boxed{
p\notin\mathcal D(N)
\Longrightarrow
\Phi_N^{\min}(p)
\notin\mathcal D(N).
}
$$

This can serve as a forbidden-color rule for obstruction coloring.

---

# 24. The Special Status of $r=2$

Since $N$ is even.

For any odd prime:

$$
p>2,
$$

the complement:

$$
N-p
$$

is odd.

Therefore:

$$
2\nmid(N-p).
$$

Thus:

$$
\Phi_N^{\min}(p)\neq2.
$$

The only prime input that may use obstruction color $2$ is:

$$
p=2.
$$

If:

$$
F_G(N)=1,
$$

then:

$$
N-2
$$

must be composite.

Since $N$ is even:

$$
N-2
$$

is even as well.

Therefore:

$$
\Phi_N^{\min}(2)=2.
$$

Thus:

$$
\boxed{
\mathcal O_2^{\min}(N)=\{2\}
}
$$

holds for a hypothetical Goldbach failure.

This is the first completely fixed cell of the canonical obstruction partition.

---

# 25. The Branch Behavior of $r=3$

For:

$$
p>3,
$$

a prime must satisfy:

$$
p\equiv1
\quad\text{or}\quad
5
\pmod6.
$$

Also, since $N$ is even:

$$
N\bmod6
\in
\{0,2,4\}.
$$

Consider:

$$
N-p\bmod6.
$$

This gives three different obstruction geometries.

---

# 26. $N\equiv0\pmod6$

If:

$$
N\equiv0\pmod6,
$$

then:

$$
p\equiv1\pmod6
\Longrightarrow
N-p\equiv5\pmod6,
$$

and:

$$
p\equiv5\pmod6
\Longrightarrow
N-p\equiv1\pmod6.
$$

Thus, for:

$$
p>3,
$$

every complement satisfies:

$$
N-p\equiv\pm1\pmod6.
$$

Therefore:

$$
2\nmid(N-p),
$$

and:

$$
3\nmid(N-p).
$$

Hence the least prime factor of every composite complement must satisfy:

$$
\boxed{
\Phi_N^{\min}(p)\ge5.
}
$$

This is the first hard-obstruction condition for a failure state with $N\equiv0\pmod6$.

---

# 27. $N\equiv2\pmod6$

If:

$$
N\equiv2\pmod6,
$$

then:

$$
p\equiv1\pmod6
\Longrightarrow
N-p\equiv1\pmod6,
$$

And:

$$
p\equiv5\pmod6
\Longrightarrow
N-p\equiv3\pmod6.
$$

Thus the second branch:

$$
p\equiv5\pmod6
$$

automatically satisfies:

$$
3\mid(N-p),
$$

unless:

$$
N-p=3,
$$

but that case directly gives a Goldbach representation, contradicting the failure assumption.

Thus, in a failure state, this branch can be absorbed extensively by $r=3$ obstruction.

In contrast:

$$
p\equiv1\pmod6
$$

forms the hard branch.

---

# 28. $N\equiv4\pmod6$

If:

$$
N\equiv4\pmod6,
$$

then:

$$
p\equiv1\pmod6
\Longrightarrow
N-p\equiv3\pmod6,
$$

And:

$$
p\equiv5\pmod6
\Longrightarrow
N-p\equiv5\pmod6.
$$

Therefore:

$$
p\equiv1\pmod6
$$

forms the $r=3$ easy branch,

And:

$$
p\equiv5\pmod6
$$

forms the hard branch.

---

# 29. M6 Easy / Hard Branch Table

For:

$$
p>3,
$$

the results can be organized as:

| $N\bmod6$ | $p\bmod6=1$ | $p\bmod6=5$ |
|---|---|---|
| $0$ | hard, $N-p\equiv5$ | hard, $N-p\equiv1$ |
| $2$ | hard, $N-p\equiv1$ | easy, $3\mid(N-p)$ |
| $4$ | easy, $3\mid(N-p)$ | hard, $N-p\equiv5$ |

where:

- easy means there is a direct $r=3$ obstruction;
- hard means the complement has escaped $2$ and $3$; if it is still composite, its least prime factor is at least $5$.

This is not a proof of Goldbach.

It only makes the first separation of obstruction burden on $M_6$.

---

# 30. M6 Hard Branch Set

Define:

$$
\mathbb H_6(N)
=
\left\{
p\in\mathbb P_N^{-}:
p>3,
\quad
\gcd(N-p,6)=1
\right\}.
$$

For:

$$
p\in\mathbb H_6(N),
$$

If:

$$
F_G(N)=1,
$$

then:

$$
\Phi_N^{\min}(p)\ge5.
$$

Thus the subsequent M6 fixed-point line genuinely needs to study not the easy branch, but:

$$
\boxed{
\mathbb H_6(N).
}
$$

---

# 31. Generalizing M6 to the Wheel Hard Branch

Let:

$$
M_k
=
\prod_{i=1}^{k}p_i
$$

be the primorial at level $k$.

Define:

$$
\mathbb H_{M_k}(N)
=
\left\{
p\in\mathbb P_N^{-}:
\gcd(N-p,M_k)=1
\right\}.
$$

If:

$$
p\in\mathbb H_{M_k}(N)
$$

and:

$$
F_G(N)=1,
$$

then:

$$
\Phi_N^{\min}(p)>p_k.
$$

Thus wheel refinement can be reinterpreted as:

> Pushing prime inputs not yet absorbed by small-prime obstruction into a deeper hard branch.

---

# 32. The Wheel Residual and Prime-Surface Insufficiency

When:

$$
k
$$

increases:

$$
\mathbb H_{M_{k+1}}(N)
\subseteq
\mathbb H_{M_k}(N).
$$

But if research merely keeps adding:

$$
p_{k+1},
$$

until every complement is caught by some prime factor, that is only:

$$
\boxed{
\text{factor-by-factor exhaustion}.
}
$$

Thus subsequent work must track:

$$
\operatorname{CompressionGain}(k)
$$

and:

$$
\operatorname{SourceNovelty}(k).
$$

If a new wheel level only adds a divisibility test, it cannot be described as a new global mechanism.

---

# 33. Obstruction Coverage Function

Define:

$$
C_N(y)
=
\left|
\mathbb P_N^{\mathrm{cov}}(y)
\right|.
$$

and the residual:

$$
R_N(y)
=
\left|
\mathbb P_N^{\mathrm{res}}(y)
\right|.
$$

We have:

$$
C_N(y)+R_N(y)
=
|\mathbb P_N^{-}|.
$$

If:

$$
F_G(N)=1,
$$

then:

$$
R_N(D_G(N))=0.
$$

Subsequent work can study:

$$
R_N(3),
\quad
R_N(5),
\quad
R_N(7),
\quad
R_N(11),
\ldots
$$

how they decay.

But these curves are, first and foremost, diagnostics, not proofs.

---

# 34. Obstruction Entropy

To compare obstruction distributions across hypothetical failure states, one can define the normalized load:

$$
\pi_r(N)
=
\frac{
L_r^{-}(N)
}{
|\mathbb P_N^{-}|
}.
$$

If:

$$
F_G(N)=1,
$$

then:

$$
\sum_r\pi_r(N)=1.
$$

Define the obstruction entropy:

$$
H_{\mathrm{obs}}(N)
=
-
\sum_r
\pi_r(N)\log\pi_r(N).
$$

This quantity can distinguish between:

- highly concentrated coverage by a few small factors;
- an obstruction burden distributed over many larger factors.

But:

$$
H_{\mathrm{obs}}
$$

is only an analytic quantity.

It does not itself provide a Goldbach proof.

---

# 35. Obstruction Energy

One can also define a quantity incorporating depth cost:

$$
E_{\mathrm{obs}}^{(\alpha)}(N)
=
\sum_{p\in\mathbb P_N^{-}}
\left(
\Phi_N^{\min}(p)
\right)^\alpha,
$$

where:

$$
\alpha>0.
$$

If obstructions are forced to use larger least prime factors, then:

$$
E_{\mathrm{obs}}^{(\alpha)}(N)
$$

increases.

This can be used to study:

> Whether certain residue / divisor structures force a hypothetical counterexample to pay a higher obstruction cost.

Again, this is a discovery metric, not a theorem in itself.

---

# 36. Canonical Failure Certificate

If some:

$$
N
$$

satisfying:

$$
F_G(N)=1,
$$

then one of the most natural finite certificates is:

$$
\Phi_N^{\min}.
$$

That is, for every:

$$
p\in\mathbb P_N^{-},
$$

supply:

$$
r_p
=
\operatorname{lpf}(N-p),
$$

and verify:

$$
r_p\mid(N-p),
$$

$$
r_p\le\sqrt{N-p}.
$$

Thus a candidate Goldbach counterexample should be accompanied by:

```text
N
prime-domain bound
p list or reproducible prime generator
complement N-p
least prime factor r_p
divisibility witness
coverage completeness
hash
independent verifier
```

This avoids treating 'the search program says there is no representation' as the only evidence.

---

# 37. Canonical Obstruction State

For a hypothetical failure state $N$, define:

$$
\mathfrak O_N
=
(
N,
\mathbb P_N^{-},
\Phi_N^{\min},
\mathcal L_N,
D_G(N),
\Theta_N,
\mathcal D(N),
\mathbb H_6(N),
\Pi_N,
C_N
).
$$

where:

- $N$: candidate failure state;
- $\mathbb P_N^{-}$: reduced prime domain;
- $\Phi_N^{\min}$: canonical obstruction map;
- $\mathcal L_N$: obstruction-load spectrum;
- $D_G(N)$: obstruction depth;
- $\Theta_N$: truncated cover curve;
- $\mathcal D(N)$: prime divisors of $N$;
- $\mathbb H_6(N)$: M6 hard branch;
- $\Pi_N$: provenance;
- $C_N$: certificates.

---

# 38. Serialization Schema

It can be stored as:

```yaml
goldbach_obstruction_state:
  version:
  N:
  domain:
    prime_lower_bound:
    prime_upper_bound:
    reduced_domain: true
  failure_status:
  obstruction_map:
    canonical: least_prime_factor
  obstruction_cells:
  obstruction_loads:
  obstruction_depth:
  primorial_cutoffs:
  covered_counts:
  residual_counts:
  divisor_suppression_set:
  m6:
    residue_of_N:
    easy_branch:
    hard_branch:
  wheel_hard_branches:
  proof_constraints:
  heuristic_metrics:
  certificates:
  provenance:
  checkpoint_hash:
```

This allows RG-SCP to extend layer by layer, rather than reconstructing prime/complement analysis in every round.

---

# 39. Three Kinds of Cover in the Research

To avoid terminological confusion, this paper distinguishes:

## 39.1 Raw Factor Cover

$$
\mathbb P_N
\subseteq
\bigcup_r
\mathcal O_r(N).
$$

May overlap.

## 39.2 Canonical Least-Factor Partition

$$
\mathbb P_N
=
\bigsqcup_r
\mathcal O_r^{\min}(N).
$$

Disjoint.

## 39.3 Truncated Wheel Cover

Allow only:

$$
r\le y.
$$

This yields:

$$
\mathbb P_N^{\mathrm{cov}}(y).
$$

It generally covers only part of the prime domain.

The three cannot be used interchangeably.

---

# 40. Three Kinds of 'Closure' in the Research

Likewise, distinguish:

## 40.1 Finite-factor closure

For fixed $N$, check all the way to:

$$
\sqrt N.
$$

This is finite verification.

## 40.2 Structural closure

Some finite or uniform structure directly implies:

$$
R_N(y)=0
$$

or another equivalent impossibility, rather than trial division term by term.

## 40.3 Global Goldbach closure

Prove:

$$
\forall N\ge4,
\quad
F_G(N)=0.
$$

Only the third kind constitutes a complete Goldbach proof.

---

# 41. A First Quantitative Version of the Cost of Counterexample Existence

The first paper proposed:

> A counterexample must pay its own cost of existence.

This paper can now give the first explicit version.

If:

$$
F_G(N)=1,
$$

then it must pay at least the following:

### C1

For every:

$$
p\in\mathbb P_N^{-}
$$

supply a composite complement.

### C2

Supply a proper prime factor for every complement.

### C3

Every obstruction factor must obey:

$$
p\equiv N\pmod r.
$$

### C4

If:

$$
r\mid N,
$$

then $r$ can cover almost no other prime inputs.

### C5

Every M6 hard-branch input must use:

$$
r\ge5.
$$

### C6

Every deeper wheel hard-branch input must use a larger factor.

This turns 'no prime pair exists' from a single negation into a large collection of simultaneous obligations.

---

# 42. Coverage Itself Must Not Be Mistaken for Proof

A special warning is needed:

For a complement already known to be composite,

finding:

$$
r\mid(N-p)
$$

only proves that it is composite.

If we do this separately for every:

$$
p
$$

we are verifying a fixed $N$.

This has not solved:

$$
\forall N.
$$

Thus:

$$
\boxed{
\text{Complete cover for one }N
\neq
\text{Goldbach theorem}.
}
$$

Even completing:

$$
\forall N\le B,
$$

over a very large finite interval still gives only finite verification.

---

# 43. Structural Questions Genuinely Worth Pursuing

Once the obstruction cover is established, genuinely valuable questions at the next layer include:

1. Do certain $N\bmod M_k$ force the hard branch to be too large?
2. Can divisor self-suppression become incompatible with prime density?
3. Is there a necessary lower bound on obstruction load?
4. Can the hard branch be transported by fixed points?
5. Is there a residual fiber that no finite wheel can explain?
6. Are some obstruction colorings impossible to realize from a single $N$?
7. Can a smaller failure state be derived from a minimal counterexample $N_*$?
8. Is there a source-class escape that recasts the factor cover as a higher-order invariant?

These are the main research directions for subsequent RG-SCP work.

---

# 44. Interface with the M6 Fixed-Point Line

The core of the previous M6 line was not:

$$
6k\pm1
$$

itself, but separating the prime-candidate sector from the fixed-point predicate.

Within RG-SCP one can further define:

$$
\mathbb H_6^{\mathrm{fix}}(N)
=
\left\{
p\in
\operatorname{Fix}(T_6):
p\le\frac N2,
\quad
\gcd(N-p,6)=1
\right\}.
$$

If:

$$
F_G(N)=1,
$$

then for every:

$$
p\in
\mathbb H_6^{\mathrm{fix}}(N)
$$

there must exist:

$$
r\ge5
$$

such that:

$$
r\mid(N-p).
$$

Thus the next stage can study:

$$
\boxed{
\text{fixed-point fiber}
\to
\text{hard obstruction cover}.
}
$$

---

# 45. Interface with Prime-Surface Insufficiency

This paper's obstruction cover makes extensive use of:

- divisibility；
- residues；
- wheel；
- least factors.

These remain very close to the prime surface / factor surface.

Thus this paper itself must also undergo a source audit.

If all future progress consists only of:

$$
2
\to
3
\to
5
\to
7
\to
11
\to
\cdots
$$

adding prime factors one by one, then RG-SCP has not obtained a new proof mechanism.

Genuine progress should involve:

$$
\boxed{
\text{many obstruction obligations}
\to
\text{one higher-order structural constraint}.
}
$$

This is one criterion for subsequent source-class escape.

---

# 46. The First Compression Target

This paper therefore proposes the first explicit compression target:

Given a hypothetical failure state $N$, seek a structure:

$$
\mathcal I(N)
$$

such that many independent:

$$
r\mid(N-p)
$$

obligations can be written uniformly as:

$$
\mathcal I(N)=0
$$

or:

$$
\mathcal I(N)\in\mathcal Z
$$

or a similar finite higher-order condition.

Only if such a:

$$
\mathcal I
$$

exists and is not a trivial recoding of the original obstruction list does it constitute genuine source compression.

---

# 47. The Second Compression Target

For the canonical obstruction map:

$$
\Phi_N^{\min},
$$

seek a proof that it cannot be arbitrary.

For example, prove that:

$$
\Phi_N^{\min}
$$

must simultaneously satisfy:

- residue coherence；
- load balance；
- divisor suppression；
- M6 hard-branch lower bounds；
- cross-wheel compatibility；
- minimal-counterexample compatibility.

If these restrictions are mutually incompatible, they may gradually compress:

$$
\mathcal A_k.
$$

---

# 48. The Core Theorem Group of This Paper

The core results that genuinely stand up at present include:

## T1

Goldbach failure is equivalent to existence of a total proper obstruction map.

$$
F_G(N)=1
\Longleftrightarrow
\exists\Phi_N.
$$

## T2

The canonical least-factor obstruction map exists and is unique in a failure state.

$$
\Phi_N^{\min}(p)
=
\operatorname{lpf}(N-p).
$$

## T3

The canonical obstruction cells partition the prime domain.

$$
\mathbb P_N
=
\bigsqcup_r
\mathcal O_r^{\min}(N).
$$

## T4

If:

$$
r\mid N,
$$

then:

$$
\mathcal O_r(N)
\subseteq
\{r\}.
$$

## T5

For:

$$
N\equiv0\pmod6
$$

and:

$$
p>3,
$$

in a failure state, the least prime factor of every complement is at least:

$$
5.
$$

These results do not themselves prove Goldbach, but supply exact interfaces for the next round's fixed-point / wheel / obstruction-burden research.

---

# 49. Falsifiability Conditions

The research value proposed in this paper must also be capable of failing.

### F1

The obstruction map supplies no new cross-$p$ constraints beyond reformulating compositeness.

### F2

The obstruction-load spectrum has no stable structure across different $N$.

### F3

Divisor self-suppression cannot be turned into any global pressure.

### F4

The M6 hard branch supplies no additional proof compression beyond standard wheel filtering.

### F5

Every higher-order invariant ultimately becomes only a deterministic readout of complete factorization.

If these situations persist, RG-SCP should stop expanding on the factor surface and turn to source-class escape.

---

# 50. Interface with the Next Paper

This paper has rewritten Goldbach failure as:

$$
\boxed{
\text{complete proper obstruction assignment}.
}
$$

The natural question for the next paper is:

> Under M6 and general wheel quotients, how is the burden of this obstruction assignment split into easy and hard branches? And does a hypothetical counterexample gradually lose available small-factor channels because of its own residue / divisor structure?

Thus the next paper enters:

# RG-SCP 03
## M6/Wheel Obstruction Geometry and Counterexample Coverage Cost
### M6/Wheel Obstruction Geometry and Counterexample Coverage Cost

The core objects will include:

$$
\mathbb H_6(N),
$$

$$
\mathbb H_{M_k}(N),
$$

$$
L_r(N),
$$

$$
D_G(N),
$$

and:

$$
\operatorname{SelfSuppression}(N).
$$

The objective is the first genuine comparison of the 'cost of existence' across different hypothetical failure states.

---

# 51. Conclusion

This paper advances the first paper's abstract:

$$
\text{counterexample admissibility}
$$

into an exact factor-obstruction structure.

To exist, a Goldbach counterexample $N$ cannot merely achieve:

> No two primes were found.

It must achieve:

$$
\forall p\in\mathbb P_N^{-},
\quad
N-p
\text{ are all composite}.
$$

Equivalently, it must supply a complete proper obstruction map:

$$
\Phi_N:
\mathbb P_N^{-}
\to
\mathbb P
$$

such that:

$$
\Phi_N(p)\mid(N-p)
$$

and:

$$
\Phi_N(p)\le\sqrt{N-p}.
$$

Therefore:

$$
\boxed{
\text{Goldbach failure}
=
\text{complete prime-indexed obstruction system}.
}
$$

And the canonical least-factor map:

$$
\Phi_N^{\min}(p)
=
\operatorname{lpf}(N-p)
$$

allows this obstruction system to be:

- partitioned;
- counted;
- ordered;
- serialized;
- projected onto M6;
- generalized to wheels;
- connected to fixed points;
- subjected to a source-class audit.

Most importantly, this paper first turns the 'cost of counterexample existence' into a computable and provable structure:

$$
\boxed{
\text{counterexample}
\Longrightarrow
\text{obstruction cover}
\Longrightarrow
\text{obstruction partition}
\Longrightarrow
\text{coverage burden}.
}
$$

This is the first step by which RG-SCP formally moves from 'reverse thinking' into 'reverse structural closure'.

---

# 52. The Shortest Version

If only one sentence is retained:

> **If a Goldbach counterexample exists, it must provide a proper prime factor for every candidate prime $p$ to block $N-p$ from being prime; thus a counterexample is equivalent to a proper obstruction system covering the entire prime-complement domain.**

Formally:

$$
\boxed{
F_G(N)=1
\Longleftrightarrow
\exists\Phi_N
\text{ such that }
\forall p\in\mathbb P_N^{-},
\quad
\Phi_N(p)\mid(N-p),
\quad
\Phi_N(p)\le\sqrt{N-p}.
}
$$

The next step is therefore not to keep finding more factors individually, but to investigate:

$$
\boxed{
\text{whether this complete obstruction system can admissibly exist as a finite higher-order structure.}
}
$$