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Proof step 70 from proof int2bytes-len-neg-01 (previous step)

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)

User applied rule cut formula

And obtained the following new goals

Proof step 71:
9 + 8 * i→n(i) = 8 * (i→n(i) + 1) + 1, 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

fma: 9 + 8 * i→n(i) = 8 * (i→n(i) + 1) + 1

Additional information

Used simplifier rules: ⊦ m * n = n * m;% ⊦ m * (n + 1) = m * n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m + n = m + n0 ↔ n = n0; Used formulas: 9 + 8 * i→n(i) = 8 * (i→n(i) + 1) + 1