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Proof step 17 from proof uint2bits-len (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
2 ^ (i0 + 1) < 2 * 2 ^ i0, 0 ≤ 2 ^ (2 + i0), 1 < 2 ^ (2 + i0), 0 < 2 ^ (2 + i0), 2 ^ (i0 + 1) < 2 ^ (2 + i0), 0 ≤ i0, 0 < j, j < 2 ^ (2 + i0), 2 ^ (i0 + 1) ≤ j, 0 ≤ j, j < 2 * 2 ^ i0, 0 < 2 ^ i0, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1) → # uint2bits(j0) = i→n(i + 2) ⊦ ~1 = i0, j = 1, # uint2bits(j / 2) + 1 = i→n(3 + i0)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)