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Proof step 40 from proof int2bytes-len-neg-01 (previous step)

0 ≤ i ∧ 0 ≤ 2, n0 ≤ i→n(i) + i→n(2), i + 1 < n→i(n0), j < 0, 0 ≤ 2 ^ (n→i(n0) * 8 - 1) + j, j + 2 ^ (n→i(n0) * 8 - 2) < 0, j + 2 ^ (8 * (i + 1) - 1) < 0, 0 < j + 2 ^ (8 * (i + 1)), 0 ≤ i ⊦ n0 = 2 + i→n(i), n0 * 8 = 1, n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
n0 < 2 + i→n(i), 0 ≤ i, 0 < j + 2 ^ (8 * (i + 1)), j + 2 ^ (8 * (i + 1) - 1) < 0, j + 2 ^ (n→i(n0) * 8 - 2) < 0, 0 ≤ 2 ^ (n→i(n0) * 8 - 1) + j, j < 0, i + 1 < n→i(n0) ⊦ n0 = 0, n0 * 8 = 1, n0 = 2 + i→n(i)

Rule argument

Additional information

Used simplifier rules: ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ m < n → n ≠ 0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ m ≠ n → (m ≤ n ↔ m < n);% ⊦ m + n = n + m;