mrwulf 8795e82865 S5.3: consRecBinding PROVEN — Theorem 3 steps 1-2 kernel-checked (the hardest object) ...

The single hardest proof in the corpus is complete, no sorry. Under the
value-equality invariant y = MTH D₁, an accepting ConsRec fold either
makes extractConsNode output a genuine collision or its first component
is the honest prefix root MTH(D₁.take n₀).

- ConsRec base changed from list-match to decidable if (if C=[] /
  if C.length=1) — same root-cause fix as Root, avoids WF-unfold
  exhaustiveness obligations; more faithful to the deployed Python.
  Whole chain (Basic..Consistency) rebuilt clean.
- consRecBinding by ConsRec.induct (10 cases): 4 base/singleton, 3
  rejection/none contradictions, 2 recursive (n₀≤k, n₀>k). The n₀>k
  none-branch is where all S5.1 infrastructure interlocks:
  kbelow_prefix_eq (prefix splits at same k) + take_take_le +
  take_drop_prefix assemble x = hnode s xx into MTH(D₁.take n₀). The
  collision branches use append_inj (fixed-width Hash) + MTH_split.
- take_all helper (take-whole-list).

Cone [propext, Classical.choice, LTLAcc.sha256, Quot.sound] — single hash
axiom. 24 certs green. Fable statement-audit: matches paper Thm 3
steps 1-2. LTL untouched (12 leaves, bcd15f9d).

Co-Authored-By: Claude Fable 5 <noreply@anthropic.com>

2026-07-11 19:18:26 +02:00

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