The full 2-adic gate reduces ten lift rows to Hensel graphs

Exact statement

Let $F(v,y)=3^aP(v)(2^x+3^y)+1$ . Eleven rows of FPRD-IC-T17 have fixed valuation $v_2(F)=N_0$ . In each of the other ten rows, the solutions of $F(v,y)\equiv0\pmod{2^e}$ form a one-dimensional Hensel graph: modulo $L_e=3\cdot2^{e-2}$ , there are exactly $2^{e-11}$ states and projection to the $v$ -coordinate is bijective.

StatusProved by LTE and complete finite base checks; independently reconstructed

External reviewNo documented external or specialist review of these FPRD results is recorded.

Context

The earlier quotient enforced the valuation selecting x, but equality also forces N to equal the 2-adic valuation of the entire left side. That second gate removes a false two-dimensional freedom.

Proof or evidence

For the ten nonrigid rows, shifting y by the current exponent period changes the next 2-adic digit because v2(3^L-1) is exact. Each v-bit therefore has a unique y-bit lift.

Full valuation gate and Hensel graph →

Verification notes

Independent C++ and Python implementations reconstruct every lift through e=22 and verify 2,048 states per row with bijective v-projection.

Limitations

The theorem concerns the 21 rows inherited from one normalized positive-type-II quotient.
A 2-adic graph alone does not exclude its states; the aligned odd-prime certificate is FPRD-IC-T21.
No external review or novelty determination is recorded.

Open work

Reuse the full-side valuation gate before applying odd-prime quotients in the remaining fork orientations.