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Proof step 34 from proof byte-law-mult-one (previous step)

0 < j0, 0 ≤ j0, 0 ≤ i0, j0 < 256, 127 < j0 ⊦ i2b(j0 * j) = i2b((j0 - 256) * j)

User applied rule cut formula

And obtained the following new goals

Proof step 35:
(j0 - 256) * j = ~ j * 256 + j * j0, 0 < j0, 0 ≤ j0, 0 ≤ i0, j0 < 256, 127 < j0 ⊦ i2b(j0 * j) = i2b((j0 - 256) * j)

Rule argument

fma: (j0 - 256) * j = ~ j * 256 + j * j0

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ ~ i * j = ~(i * j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + ~ j = i - j;% ⊦ i * j - i * j0 = i * (j - j0);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used formulas: (j0 - 256) * j = ~ j * 256 + j * j0