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Proof step 23 from proof int2bytes-len-neg-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 24:
(n→i(n0) - 2) + 1 = n→i(n0) - 1, 0 < j + 2 ^ (8 * ((n→i(n0) - 2) + 1)), j + 2 ^ (8 * (i + 1)) < j + 2 ^ (8 * ((n→i(n0) - 2) + 1)), i + 2 < n→i(n0), 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 ⊦ n0 = 0, n0 * 8 = 1, n0 = 2 + i→n(i)

Rule argument

fma: (n→i(n0) - 2) + 1 = n→i(n0) - 1

Additional information

Used simplifier rules: ⊦ i + (j - i0) = i + j - i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + (j - i) = j;% ⊦ n→i(m) = n→i(n) ↔ m = n; Used formulas: (n→i(n0) - 2) + 1 = n→i(n0) - 1