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Proof step 9 from proof uint2bits-len-01 (previous step)

0 ≤ i, j < 3 * 2 ^ i, 2 * 2 ^ i ≤ j, 0 < j, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ # uint2bits(j / 2) + 1 = i→n(i + 3), j = 1, 0 ≠ i ∧ ¬ 3 * 2 ^ (i - 1) ≤ j / 2 ∧ ¬ j / 2 < 2 * 2 ^ (i - 1) ∧ 0 ≠ i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
i = i0 + 1, 0 ≤ i, j < 3 * 2 ^ i, 2 * 2 ^ i ≤ j, 0 < j, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ # uint2bits(j / 2) + 1 = i→n(i + 3), j = 1, 0 ≠ i ∧ ¬ 3 * 2 ^ i0 ≤ j / 2 ∧ ¬ j / 2 < 2 * 2 ^ i0 ∧ 0 ≠ i

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 1, i0] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i0