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Proof step 15 from proof int2bytes-len-pos (previous step)

8 ≠ 0, ∃ n0. # int2bits(j) = n0 * 8, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# int2bits(j) = n * 8, 8 * i < abs(8 * (i + 1) - 1), 0 < j, i→n(8 * 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

Rule argument

Additional information

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