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theorem proved term proof high

ratio_only

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A zero-parameter ledger forces all level ratios to be powers of the single adjacent ratio. Researchers closing the locality gap in the Recognition Science forcing chain cite this result. The proof relies on induction over the natural-number indices combined with the constant-ratio axiom and linear-arithmetic closure.

claimFor a zero-parameter ledger $L$ whose level values are strictly positive and whose adjacent ratios are constant, the ratio of levels at indices $m$ and $n$ equals the base adjacent ratio raised to the power $m-n$.

background

The ZeroParameterLedger structure encodes a ledger with no external scales: its level sequence satisfies a constant adjacent ratio independent of any positive multiplier. This property is the key hypothesis for deriving dimensionless composition rules. The surrounding module uses this to bridge from ledger axioms to the local binary recurrence required by the T5 to T6 step in the unified forcing chain.

proof idea

Proof by induction on the first index m. When m is zero a nested induction on n applies the no-external-scale property to equate ratios, rewrites via division equivalence, and invokes nlinarith on the positivity facts. The successor case for m similarly invokes the scale-invariance axiom once and closes the equality with nlinarith using all positivity assumptions.

why it matters in Recognition Science

This result is the initial lemma in the derivation that zero parameters imply ratio-only composition, which then supports the binary-is-minimal theorem and ultimately the locality-from-ledger statement. It directly addresses the gap-1b question in the module by showing that composition rules must be expressible through adjacent-level ratios alone. Within the broader framework it contributes to the T5-J-uniqueness to T6-phi-fixed-point transition by enforcing locality without additional parameters.

scope and limits

formal statement (Lean)

  52theorem ratio_only (L : ZeroParameterLedger) :
  53    ∀ m n : ℕ, L.levels m / L.levels n =
  54      (L.levels 0 / L.levels 1) ^ ((m : ℤ) - n) := by

proof body

Term-mode proof.

  55  intro m n
  56  induction m with
  57  | zero =>
  58    induction n with
  59    | zero => simp
  60    | succ n ih =>
  61      have h := L.no_external_scale 1 one_pos n
  62      rw [div_eq_div_iff (ne_of_gt (L.all_positive 0)) (ne_of_gt (L.all_positive (n + 1)))]
  63      have := L.no_external_scale 1 one_pos 0
  64      rw [Nat.zero_add] at this
  65      rw [div_eq_div_iff (ne_of_gt (L.all_positive 0)) (ne_of_gt (L.all_positive 1))] at this
  66      nlinarith [L.all_positive 0, L.all_positive 1, L.all_positive n, L.all_positive (n+1)]
  67  | succ m ih =>
  68    have h := L.no_external_scale 1 one_pos m
  69    rw [div_eq_div_iff (ne_of_gt (L.all_positive (m+1))) (ne_of_gt (L.all_positive n))]
  70    have hm := L.no_external_scale 1 one_pos 0
  71    nlinarith [L.all_positive m, L.all_positive (m+1), L.all_positive n, L.all_positive 0, L.all_positive 1]
  72

depends on (6)

Lean names referenced from this declaration's body.