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Proof step 8 from proof uint2bits-len (previous step)

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

Heuristic var elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
i = i0 + 1, 2 ^ i ≤ j, j < 2 ^ (i + 1), 0 < j, 0 < 2 ^ (i + 1), 0 ≤ 2 ^ (i + 1), 1 < 2 ^ (i + 1), 0 < i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1) → # uint2bits(j0) = i→n(i + 2) ⊦ # uint2bits(j / 2) + 1 = i→n(i + 2), j = 1, 0 = i, ¬ 2 ^ i ≤ j / 2 ∧ ¬ j / 2 < 2 ^ i0

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