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Proof step 5 from proof byte-law-mult-one (previous step)

~256 < j0, i0 ≤ 0, i0 * 256 + j0 < 0, j0 < 0 ⊦ j0 = 0, i2b((i0 * 256 + j0) * j) = i2b(b2i(i2b(j0)) * j)

User applied rule cut formula

And obtained the following new goals

Proof step 6:
(i0 * 256 + j0) * j = (i0 * j) * 256 + j0 * j, ~256 < j0, i0 ≤ 0, i0 * 256 + j0 < 0, j0 < 0 ⊦ j0 = 0, i2b((i0 * 256 + j0) * j) = i2b(b2i(i2b(j0)) * j)

Rule argument

fma: (i0 * 256 + j0) * j = (i0 * j) * 256 + j0 * j

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < 0 → (i * j = i * j0 ↔ j = j0); Used formulas: (i0 * 256 + j0) * j = (i0 * j) * 256 + j0 * j