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Proof step 12 from proof uint2bits-len-01 (previous step)

j = i1 * 2 + j1 ∧ 0 ≤ j1 ∧ j1 < 2 ∧ 0 < 2 ∧ 0 ≤ j, ~1 ≤ i0, 0 < j, 2 * 2 ^ (i0 + 1) ≤ j, j < 3 * 2 ^ (i0 + 1), 0 < 2 ^ (i0 + 1), ∀ 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(i1) + 1 = i→n(4 + i0), j = 1, ~1 ≠ i0 ∧ ~1 ≠ i0 ∧ ¬ 3 * 2 ^ i0 ≤ i1 ∧ ¬ i1 < 2 * 2 ^ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
~1 ≤ i0, 0 < i1 * 2 + j1, 2 * 2 ^ (i0 + 1) ≤ i1 * 2 + j1, i1 * 2 + j1 < 3 * 2 ^ (i0 + 1), 0 < 2 ^ (i0 + 1), 0 ≤ j1, j1 < 2, 0 ≤ i1, ∀ j0, i. i→n(j0) < i→n(i1 * 2 + j1) → 0 ≤ i ∧ j0 < 3 * 2 ^ i ∧ 2 * 2 ^ i ≤ j0 ∧ 0 < 2 ^ i → # uint2bits(j0) = i→n(i + 3) ⊦ # uint2bits(i1) + 1 = i→n(4 + i0), i1 * 2 + j1 = 1, ~1 ≠ i0 ∧ ~1 ≠ i0 ∧ ¬ 3 * 2 ^ i0 ≤ i1 ∧ ¬ i1 < 2 * 2 ^ i0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);