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Proof step 33 from proof int2bytes-len-neg (previous step)

j < 0 ∧ 0 ≤ 8 * i ∧ ~ j ≠ 2 ^ (8 * i), ¬ # int2bits(j) ≤ i→n(8 * i) + 1, 0 ≤ 2 ^ (8 * (i + 1) - 1) + j, 0 ≤ i, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1) ⊦ # int2bits(j) % 8 = 0, # int2bits(j) / 8 = i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
j < 0, i→n(8 * i) + 1 < # int2bits(j), 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1), 0 ≤ i, 0 ≤ 2 ^ (8 * (i + 1) - 1) + j ⊦ j + 2 ^ (8 * i) = 0, # int2bits(j) / 8 = i→n(i), # int2bits(j) % 8 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ ~ i = j ↔ i + j = 0;