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Proof step 28 from proof int2bytes-rec-neg-03 (previous step)

0 < i, j + 2 ^ (i * 8) < 0 ⊦ j % 2 ^ (i * 8) = 0, int2bytes(j / 2 ^ (i * 8) - 1) = int2bytes((j + 1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

User applied rule cut formula

And obtained the following new goals

Proof step 29:
j + 1 - 2 ^ (i * 8) = (j + 1) + ~1 * 2 ^ (i * 8), 0 < i, j + 2 ^ (i * 8) < 0 ⊦ j % 2 ^ (i * 8) = 0, int2bytes(j / 2 ^ (i * 8) - 1) = int2bytes((j + 1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

Rule argument

fma: j + 1 - 2 ^ (i * 8) = (j + 1) + ~1 * 2 ^ (i * 8)

Additional information

Used simplifier rules: ⊦ i * ~1 = ~ i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + ~ j = i - j;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + (j - i) = j;% ⊦ i + j = i + i0 ↔ j = i0; Used formulas: j + 1 - 2 ^ (i * 8) = (j + 1) + ~1 * 2 ^ (i * 8)