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Proof step 69 from proof int2bytes-len-neg-01 (previous step)

0 ≤ i ∧ 0 ≤ 2, i < n→i(m), 0 ≤ n→i(m) * 8 + n→i(n), 0 < n→i(m) * 8 + n→i(n), 2 ≤ n→i(m) * 8 + n→i(n), m * 8 + n ≤ 9 + 8 * i→n(i), j + 2 ^ (8 * (i + 1) - 1) < 0, j < 0, n < 8, 8 * (i + 1) < n→i(m) * 8 + n→i(n), 0 ≤ 2 ^ (n→i(m) * 8 + n→i(n) - 1) + j, j + 2 ^ (n→i(m) * 8 + n→i(n) - 2) < 0, 0 ≤ i ⊦ m = 0, n = 0, m + 1 = i→n(i) + i→n(2), j + 2 ^ (8 * (i + 1)) = 0, m * 8 + n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
i < n→i(m), 0 ≤ n→i(m) * 8 + n→i(n), 0 < n→i(m) * 8 + n→i(n), 2 ≤ n→i(m) * 8 + n→i(n), m * 8 + n ≤ 9 + 8 * i→n(i), j + 2 ^ (8 * (i + 1) - 1) < 0, j < 0, n < 8, 8 * (i + 1) < n→i(m) * 8 + n→i(n), 0 ≤ 2 ^ (n→i(m) * 8 + n→i(n) - 1) + j, j + 2 ^ (n→i(m) * 8 + n→i(n) - 2) < 0, 0 ≤ i ⊦ j + 2 ^ (8 * (i + 1)) = 0, m * 8 + n = 0, m = 0, n = 0, m = 1 + i→n(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j + i0 = (i + j) + i0;% ⊦ m * n = n * m;% ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n);