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Proof step 60 from proof int2bytes-len-neg (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 61:
8 + 8 * i→n(i) = 8 * (1 + i→n(i)), 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

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

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: 8 + 8 * i→n(i) = 8 * (1 + i→n(i))