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Proof step 59 from proof int2bytes-len-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
i→n(i) < m, 0 ≤ i, j < ~1, n < 8, 0 ≤ 2 ^ (n→i(m) * 8 + n→i(n) - 1) + j, j + 2 ^ (n→i(m) * 8 + n→i(n) - 2) < 0, 2 ^ (n→i(m) * 8 + n→i(n) - 2) < 2 ^ (n→i(m) * 8 + n→i(n) - 1), j < 0, m * 8 + n ≤ 8 + 8 * i→n(i) ⊦ 8 * (i + 1) = 1, n = 0, m = i→n(i), j + 2 ^ (8 * i) = 0, ~1 = i

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n → n ≠ 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → m0 * n + n1 < m * n0 + n2;% ⊦ m < n → (n ≤ m ↔ false);% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m ≠ n → (¬ m < n ↔ n < m);% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ m * n = n * m;