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Proof step 18 from proof int2bytes-len-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# int2bits(j) = n * 8, i→n(i) + 1 < n, i→n(i) < n, 0 < j, 8 * i < abs(8 * (i + 1) - 1), 8 * i + 1 < 8 * (i + 1), n * 8 ≤ i→n(8 * (i + 1) - 1) + 1, 0 ≤ i ⊦ n = i→n(i) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward)% ⊦ m < n → n ≠ 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ n0 < m0 → (m0 * m + n0 < m0 * n ↔ m < n);