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Proof step 12 from proof int2bytes-len-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# int2bits(j) % 8 = 0, 8 * i + 1 < 8 * (i + 1), # int2bits(j) ≤ i→n(8 * (i + 1) - 1) + 1, 0 ≤ i, i→n(8 * i) + 1 < # int2bits(j), 0 < j, 8 * i < abs(8 * (i + 1) - 1) ⊦ # int2bits(j) / 8 = i→n(i) + 1

Rule argument

Additional information

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