Unicode • algebra only • evaluate exactly as written
Rosetta LT — Instantiated Final Kernel (Unicode, algebra only)
§6 tokens (Unicode)
Arrow (scalar reading): Y′ = Y (or, more generally, Y′ = a·Y with fixed a > 0).
Mixed line: Y′ = β · ( Y − λ·N ), with β > 1 and 0 < |λ| < 1 (so β, λ ≠ 0).
A) Instantiated kernel (fastest)
Take the arrow as Y′ = Y, then eliminate Y′:
Identity demanded: Y = β ( Y − λ·N ) for the same Y, N.
Pick a single concrete, non-degenerate point: Y = 0, N = 1.
This yields: 0 = β ( 0 − λ·1 ) = −βλ, impossible because βλ ≠ 0.
→ Contradiction (so the arrow and the mixed line cannot both be enforced as general identities).
B) Quantified form (for clarity)
Assume ∀Y ∀N : Y = β ( Y − λ·N ) (this is what “general identity” means).
Instantiate Y = 0, N = 1 ⇒ 0 = −βλ ⇒ impossible (β, λ ≠ 0).
Hence ∀Y ∀N cannot be true; the set is inconsistent.
C) If someone prefers the “scalar covariance” version
Let the arrow be Y′ = a·Y (fixed a > 0). Combine with the mixed line:
a·Y = β·Y − β·λ·N ⇒ (a − β)·Y = −β·λ·N.
As a general identity (for all Y, N), this is impossible because the left side scales with Y only, the right side with N only. A single scalar a cannot satisfy it for all pairs (Y, N) unless one of β or λ is zero (ruled out).
Again, inconsistency.
One-line takeaway for §6
If the §6 arrow (whether Y′ = Y or Y′ = a·Y) is taken as a general identity alongside the mixed relation Y′ = β ( Y − λN ) with β > 1 and 0 < |λ| < 1, then §6 is internally inconsistent. The contradiction appears immediately upon the concrete instantiation Y = 0, N = 1.
What the kernel is trying to do
You want a tiny algebra puzzle that anyone must judge inconsistent using only arithmetic. No talk of frames, coordinates, or context. If this mini-system is inconsistent, then any printed set of equations that matches its tokens is also inconsistent unless one of those equations is explicitly withdrawn or restricted.
The final kernel, simplified
We fix specific nonzero values so there is no wiggle room:
Assumptions:
m ≠ 0, ε ≠ 0, β > 1, and 0 < |λ| < 1 (so β and λ are both nonzero).
Three simultaneous equalities:
aₓ = (ε/m)·1
aₓ = (ε/(m·β³))·1
0 = β·(0 − λ·1)
There are no variables to choose and no quantifiers to hide behind. The task is simply: can all three equalities be true at once under the assumptions?
Why it must be inconsistent
Part A — the aₓ pair
From (1) and (2):
(ε/m)·1 = (ε/(m·β³))·1.
Because m ≠ 0 and ε ≠ 0, divide both sides by (ε/m):
1 = 1/β³ ⇒ β³ = 1 ⇒ β = 1.
But the assumptions require β > 1.
→ Contradiction 1: the system demands β = 1 and β > 1 at the same time.
Part B — the “mixed” identity
From (3):
0 = β·(0 − λ·1) = −βλ.
Since β > 1 and 0 < |λ| < 1, both β and λ are nonzero, so βλ ≠ 0.
Thus −βλ ≠ 0, but line (3) says it equals 0.
→ Contradiction 2: a nonzero number cannot equal 0.
Either contradiction alone makes the system impossible; having both is overkill.
Therefore the mini-system is INCONSISTENT.
Why earlier versions sometimes returned “CONSISTENT”
Some engines used loopholes like:
Setting X = 0, which makes two aₓ formulas trivially match.
Picking special Y, N values to make a mixed identity hold for that one case.
Claiming the equations apply in different “scopes” or “limits.”
The final kernel closes those doors by:
Fixing the multipliers to 1, so you cannot zero them out.
Forcing nonzero β and nonzero λ with the assumptions β > 1 and 0 < |λ| < 1.
Avoiding any language about frames, coordinates, or interpretation.
How this mirrors the earlier §10 / §6 tensions (optional context)
The two aₓ lines condense the §10 duplication: two different formulas for the same component aₓ in one place. Arithmetic says they can agree only if β = 1, which contradicts the condition β > 1.
The third line mimics the §6 issue: it forces a cross-term to be zero even though both factors are nonzero by assumption.
You do not have to accept any “mapping” to agree the kernel is inconsistent. The kernel stands by itself. But if a printed set of relations matches these forms simultaneously without disclaimers, the same algebraic contradictions reappear.
Short FAQ
Isn’t this “engineering a contradiction”?
It is a satisfiability test. If a theory (or a printed set of equations) includes these forms at the same time, those forms are mutually unsatisfiable unless you explicitly restrict or remove one of them.
Can we rescue it by saying one equation holds only “in a limit” or “in another context”?
That is exactly a withdrawal or scope restriction. The point is to make such restrictions explicit. Without them, the bare algebra clashes.
Why choose constants 1, β > 1, and 0 < |λ| < 1?
To block trivial escapes (like choosing 0) and to keep the arithmetic transparent.
One-paragraph takeaway
The instantiated kernel is a tiny, fully explicit algebra puzzle: assume nonzero constants and ask if three simple equalities can all be true together. They cannot—two separate contradictions appear immediately. This is pure arithmetic, not interpretation. If a set of “laws” matches these forms at the same time, then—unless someone limits or removes one of them—that set is algebraically inconsistent
These two are clear why they should be valid simultaneouslyL1) aₓ = (ε/m)·1 (2) aₓ = (ε/(m·β³))·1 but where did this come from, (3) 0 = β·(0 − λ·1), and why it should be simultaneously valid?
Great question. Short answer: line (3) is just the instantiation of the §6 “mixed” relation, and it’s included to test simultaneous validity of the two printed families (§10 and §6) in one tiny system.
Here’s the logic:
Where (3) comes from
The §6 pair is usually written (tokenwise) as
• arrow (scalar): Y′ = Y (or Y′ = a·Y, a > 0)
• mixed line: Y′ = β · ( Y − (v/c) · N )
If you eliminate Y′ (take Y′ = Y), you get the identity:
Y = β · ( Y − (v/c) · N )
Rename λ := v/c (so 0 < |λ| < 1) and pick a particular case of Y and N to avoid trivial escapes. The simplest is Y = 0 and N = 1. Substituting gives:
0 = β · ( 0 − λ · 1 ), which is exactly line (3).
So (3) is nothing mysterious; it’s just the §6 identity Y = β(Y − λN) evaluated at one concrete, non-degenerate point (Y = 0, N = 1) to make the algebra check immediate.
Why it should be valid simultaneously with (1) and (2)
The whole purpose of the kernel is a consistency test: we place in one set the token-forms that are being claimed together (the §10 pair for aₓ and the §6 arrow+mixed line for Y,N).
If the §6 statements are intended as general identities (i.e., they are printed as holding without special conditions), then they must hold for all Y and N—hence also for the specific choice Y = 0, N = 1. That single instantiation gives (3).
Likewise, (1) and (2) are the two §10 formulas for the same aₓ; in the kernel we assert them together to see if they can both be true under β > 1.
If someone says “(3) shouldn’t be enforced,” they’re implicitly withdrawing the general-identity reading of §6 (i.e., denying Y′ = Y as a scalar arrow or denying that the mixed line applies simultaneously). That’s fine—but it’s a scope restriction. The kernel’s job is to make such a withdrawal explicit; otherwise, taken together, the bare algebra clashes.
If you prefer a quantified path first
Start with: ∀Y ∀N : Y = β (Y − λN) (the general §6 identity).
Then instantiate with Y = 0, N = 1 to get 0 = −βλ.
With β > 1 and 0 < |λ| < 1, that’s impossible—hence inconsistency even before considering (1)–(2).
So, in summary:
(1) and (2) encode the §10 duplication for aₓ.
(3) encodes the §6 arrow-vs-mixed relation at one concrete (Y,N), which must be valid if the §6 statement is genuinely general.
Putting them together is intentional: it tests whether the combined printed forms can coexist without hidden scope restrictions. If not, you either (a) withdraw a statement (state the restriction), or (b) accept that the set is algebraically inconsistent.
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