SNOL: reduction to the rank-1 neighbour¶
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In 17 Payment we sharpened "the engine does not stop for free" into a precise defect law: a free pass to a prime \(p\) requires that \(a-\theta\) be divisible by the entire primorial of the small primes, and as soon as the primorial outgrows the active divisor, such a pass is impossible for a nontrivial \(a\).
But there the single distributional input — that is, a named missing statement, a gate (see the glossary) — remained isolated in a scalar: the balance of shifted-charge and tax, pulling in opposite directions and refusing to close simultaneously without counting. The present chapter changes the angle of attack: instead of trying to defeat that scalar budget by counting, we reduce the entire programme to a single algebraic node and show why that node is non-counting by construction.
Change of strategy: from the counting wall to a single node¶
All the previous final routes — four-corner (12, 14) and payment budget (17) — ran into one and the same counting wall: to close them, one had to control the distribution of the shift's divisors (Mertens, Brun/Selberg), that is, to cross the parity barrier. Intuition suggests trying not to break through this wall, but to walk around it structurally.
The observation underlying the reduction: the defects of a product-state are organised by rank (the number of primes \(>A\) in the factorisation of the active side), and along the rank one can descend. Rank descent reduces every defect to rank-1 — a state in which one side of the wedge is the active prime \(a\) itself. And rank-1, as we show below by pure algebra, reduces to one neighbour lemma:
We call this implication SNOL (Shifted-Neighbour Obstruction Lemma). The key property of SNOL — the very point of undertaking the whole reduction — is that it is non-counting by construction. Below we prove all the surrounding algebra machine-checked and present numerical evidence that counting is powerless here in both directions.
Note. Nowhere do we pass this reduction off as a proof of the twin conjecture. What is proven is the algebra of rank-1 and the conditional theorem "SNOL \(\Rightarrow\) infinitude of twins". SNOL itself is the single open node; the chapter honestly localises it — it does not close it.
The provable rank-1 algebra¶
Carrying the two across to the opposite side¶
Introduce the rank-1 configuration: the active prime \(a\) occupies one side of the wedge with centre \(n\) and sign \(\varepsilon\in\{-1,+1\}\), that is,
Then the opposite side of the same centre is computed with nothing left over.
Theorem 18.1 (
rank1_opposite). If \(6n+\varepsilon=a\), then\[ 6n-\varepsilon = a - 2\varepsilon. \]
The proof is pure carry-the-two arithmetic (omega): subtracting \(2\varepsilon\) from both sides of the equality \(6n+\varepsilon=a\), we obtain \(6n-\varepsilon=a-2\varepsilon\). Substantively this means: as soon as one side of a rank-1 wedge is the active prime, the second side is rigidly fixed as the shifted neighbour \(a-2\varepsilon\). It is precisely this shifted neighbour that becomes the arena of all the ensuing struggle — not \(a\) itself, but its neighbour \(a-2\varepsilon\) (for \(\varepsilon=+1\) this is \(a-2\), for \(\varepsilon=-1\) it is \(a+2\)).
Saturation of the neighbour corridor¶
Now suppose the active prime lies in the neighbour corridor with respect to a finite set \(Q\) of pairwise coprime small moduli: for every \(q\in Q\) we have \(a\equiv 2\varepsilon\pmod q\), equivalently \(q\mid a-2\varepsilon\). Then the divisibility lifts to the primorial.
Theorem 18.2 (
neighbour_saturation). Suppose \((Q)_{\mathrm{Set}}\) are pairwise coprime and \(q\mid a-2\varepsilon\) for all \(q\in Q\). Then\[ \Bigl(\prod_{q\in Q} q\Bigr)\ \Big|\ a-2\varepsilon, \]that is, \(a = k\!\cdot\!\prod_{q\in Q} q + 2\varepsilon\) for some \(k\).
The proof lifts the pairwise divisibility to divisibility by the product via pairwise coprimality (Finset.prod_dvd_of_coprime, with coprimality translated into IsCoprime over \(\mathbb Z\)). This is exactly the Euclidean twin-neighbour shift of the form \(kP\pm 2\): an active prime whose entire neighbour has been "eaten up" by the small primes is forced to have the form \(a=kP+2\varepsilon\). There is no distribution anywhere here — this is a divisibility identity.
The threshold: the corridor is out of reach for small \(a\)¶
The saturation \(P\mid a-2\varepsilon\) comes into conflict with size as soon as the corridor's primorial outgrows the magnitude of the shift.
Theorem 18.3 (
neighbour_corridor_bound). If \(P\mid a-2\varepsilon\) and \(|a-2\varepsilon|<P\), then \(a=2\varepsilon\).
The proof splits into two cases. If \(a-2\varepsilon=0\), the claim is trivial. Otherwise \(a-2\varepsilon\neq 0\), and from \(P\mid a-2\varepsilon\) we get \(P\le|a-2\varepsilon|\) (Int.le_of_dvd applied to a nonzero dividend) — contradicting \(|a-2\varepsilon|<P\), so the remaining case is closed by omega.
Substantively: a nontrivial active prime cannot saturate an entire corridor whose primorial has exceeded its shift. This is exactly the same defect law as in 17 Payment (shifted_primorial_bound, late_boundary_not_free), but for the neighbour shift \(2\varepsilon\) instead of the compatibility shift \(\theta\).
Note. The three theorems above — the whole of the rank-1 algebra — are proven without a single distributional assumption. Numerically the rank descent is short: the old-free composite sides \(6m\pm1\) exhibit 100% rank 2 (exactly two primes \(>A\)), so in practice the chain \(4\to3\to2\to1\) degenerates into a single step \(2\to1\); and the neighbour corridor \(kP\pm2\) is confirmed by an exact identity (\(Q=\{5,7,11\}\), \(P=385\): \(a=385k\pm2\Rightarrow a\mp2\equiv0\pmod{385}\)).
The final conditional theorem¶
To connect the rank-1 algebra with the twin conjecture, we fix the SNOL input — the same typed block predicate that closes the whole programme in 15 ToTwins.
Definition 18.4 (
SNOLInput). The statement\[ \forall N,\ \exists\,\text{carrier},\text{bad}:\ (\forall m\in\text{carrier},\ N<m)\ \wedge\ |\text{bad}|<|\text{carrier}|\ \wedge\ (\forall m\in\text{carrier},\ m\notin\text{bad}\Rightarrow \mathrm{IsTwinCenter}\,m). \]
That is: at every scale \(N\) there exists a carrier above \(N\) in which the "bad" elements (carriers of the terminal shifted-neighbour obstruction) are strictly fewer than the whole, and every survivor is a twin centre. SNOL as a substantive claim says exactly this: the terminal shifted-neighbour current cannot be carrier-scale, so the block input holds.
Theorem 18.5 (
twin_primes_of_SNOL). IfSNOLInputholds, thenTwinLowers.Infinite— there are infinitely many twin primes.
The proof is the direct bridge twin_prime_conjecture_of_blocks: the same capstone passage "block dominance of the survivors \(\Rightarrow\) unboundedness of twin centres \(\Rightarrow\) infinitude" as in 15 ToTwins, only with the input supplied in the form of SNOL rather than the genuine four-corner. Contraposition completes the picture.
Theorem 18.6 (
finite_contradicts_SNOL). IfTwinLowersis finite andSNOLInputholds, thenFalse.
Proof: finite_contradicts_SNOL hfin H := hfin (twin_primes_of_SNOL H). The single open node of the whole programme is thus SNOL itself: everything else is machine-verified.
Why SNOL is not the same wall¶
Here lies the fundamental difference from the previous inputs. Earlier the input \(H\) (four-corner) was a distribution in disguise: it could not be checked without estimating the density of ranks. SNOL is built differently, and the numbers confirm it.
The neighbour audit (tools/RESULTS_snol.md, harness snol_harness.py) measures the fraction of primes \(a\in(A,A^2]\) whose neighbour \(a\pm2\) already has a small divisor \(\le A\) — that is, the fraction where CRT "catches" the neighbour:
| \(A\) | \(a-2\) caught | \(a+2\) caught |
|---|---|---|
| 50 | 0.617 | 0.628 |
| 200 | 0.705 | 0.706 |
| 1000 | 0.778 | 0.778 |
The fraction is large and grows with \(A\) following the law \(1-\prod_{q\le A}(1-1/q)\). Hence a two-sided conclusion:
- SNOL cannot be proven by counting/CRT capacity. For 62–78% of active primes the neighbour is already caught by a small prime — this is not an anomaly but the norm. So the "catch \(p\mid a-2\varepsilon\)" is in itself typical, and no capacity estimate will forbid it.
- Consequently, SNOL is bound to rest not on distribution but on the Euclidean pedigree of the active \(a\): on the fact that \(a\) arrived from the descent of a wedge centre, rather than being drawn from the general population of primes.
Conclusion. The route does not cross the red line precisely because where the previous inputs secretly summoned distribution, SNOL forbids it (counting is powerless in both directions) and demands the structure of the pedigree in its place.
Conjecture (SNOL, open). The terminal shifted-neighbour obstruction cannot be carrier-scale: at every scale, the fraction of active primes \(a\) (arrived via descent) whose neighbour \(a-2\varepsilon\) is systematically caught by an old prime \(p\le A\) in such a way that \(a\) becomes a terminal, is strictly smaller than the carrier — that is, the survivors are the majority.
Closure plan. Tie the active prime \(a\) to its descent ancestor so that the structure of the pedigree forbids systematic landing in the neighbour corridor \(p\mid a-2\varepsilon\). Since counting is powerless here by construction, the argument must be about the provenance of \(a\), not about its arithmetic modulo the small primes. The first step of this plan is to show that the "catch" \(p\mid a-2\varepsilon\) is not a terminal but a descent step, regenerating a smaller centre.
Where we are now¶
The entire programme has been reduced to one machine-pinned node — SNOL. Unlike the previous inputs, it is non-counting (we checked: counting is powerless both for and against) and points to the exact site of future work — the pedigree of the active prime.
It is natural to expect that the final step of the closure plan — unfolding the "catch" into descent — must likewise rest not on distribution but on the already proven impossibility of Euclid's engine — the prohibition of infinite strictly decreasing descent (see the glossary). It is precisely this unfolding that we take apart next: in [19 Old-peel] the terminal shifted-neighbour obstruction ceases to be a terminal and turns out to be yet another step of strict descent, closing the finale onto EPMI without a single counting argument.