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Proof step 8 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) ⊦ 0 ≤ i - 1 ∧ j / 2 < 3 * 2 ^ (i - 1) ∧ 2 * 2 ^ (i - 1) ≤ j / 2 ∧ 0 < 2 ^ (i - 1), # uint2bits(j / 2) + 1 = i→n(i + 3), j = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
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

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 < i → (0 < i ^ j ↔ i = 1 ∨ 0 ≤ j);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;