Rosetta LT — Instantiated Final Kernel

Unicode • algebra only • evaluate exactly as written
The same “instantiate-and-check” logic applies to §10 once you note that the fourth system there uses the same field symbols X,Y,N X,Y,N that are already duplicitous in §6: From §6 you have the arrow (scalar) reading: Y′=Y Y′=Y (or Y′=a⋅Y Y′=a⋅Y, fixed a>0 a>0). In the same §6, you also have the mixed line: Y′=β (Y−(v/c) N) Y′=β(Y−(v/c)N). If you enforce both as general statements, you are forced to accept the identity Y=β (Y−λ N)with λ:=v/c,  0<∣λ∣<1,  β>1. Y=β(Y−λN)with λ:=v/c,0<∣λ∣<1,β>1. Instantiate at a single non-degenerate point, say Y=0,  N=1 Y=0,N=1. Then 0=β (0−λ⋅1)=−βλ, 0=β(0−λ⋅1)=−βλ, which is impossible because βλ≠0 βλ  =0. So the §6 pair is algebraically incompatible unless you withdraw one of them. Now look at §10. Its fourth system writes the dynamics in the same frame K K using the same X,Y,N X,Y,N. Two things now happen: The §10 x–acceleration duplication (single frame K K, same X X): ax=(ε/m) Xandax=(ε/(m β3)) X. a x ​ =(ε/m)Xanda x ​ =(ε/(mβ 3 ))X. For X≠0 X  =0 these are equal only if β=1 β=1, contradicting 0<∣v∣1 0<∣v∣1. → §10 already conflicts unless you withdraw one of these two laws. The §6 contradiction propagates into §10’s fourth system: That system assumes the field components Y Y and N N are available as independent inputs on the same footing as X X. But if you also keep both §6 statements, you inherit the impossible identity Y=β(Y−λN) Y=β(Y−λN). The moment you evaluate at any single pair like (Y,N)=(0,1) (Y,N)=(0,1), you force 0=−βλ 0=−βλ, which cannot be satisfied under β>1 β>1 and 0<∣λ∣<1 0<∣λ∣<1. → The field side of §10’s fourth system becomes unsatisfiable unless you withdraw one of the §6 statements. Putting these together: with the same symbols X,Y,N X,Y,N used across sections, the combined printed set { {§10 fourth-system laws, §10 x-acceleration pair, §6 arrow, §6 mixed line } } is algebraically inconsistent unless at least one of the offending lines is explicitly withdrawn or scope-restricted. That’s the §10 analogue of the §6 “duplicitous fields” contradiction. If you want the minimal, concrete kernel that shows the combined (§10+§6) clash in three lines (no quantifiers, full Unicode), use this:

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