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Proof step 9 from proof int2bytes-len-neg-01 (previous step)

8 ≠ 0, ∃ n0. # int2bits(j) = n0 * 8, ∃ n0. # int2bits(j) = n0 * 8, 0 ≤ i, 0 < j + 2 ^ (8 * (i + 1)), j + 2 ^ (8 * (i + 1) - 1) < 0 ⊦ # int2bits(j) / 8 = i→n(i + 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# int2bits(j) = n0 * 8, n * 8 = n0 * 8, j + 2 ^ (8 * (i + 1) - 1) < 0, 0 < j + 2 ^ (8 * (i + 1)), 0 ≤ i ⊦ n0 = i→n(i + 2)

Rule argument

Additional information

Used simplifier rules: ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;%n ≠ 0 ⊦ (m * n) / n = m;