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Proof step 15 from proof uint2bits-len-01 (previous step)

¬ 0 ≤ i0, ~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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i < i + j ↔ 0 < j;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0;% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i < 0 → 2 ^ i = 0;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ ¬ i ≤ j ↔ j < i;