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Proof step 40 from proof byte-mult-256 (previous step)

i0 < 0, 0 < i, ~256 < j0, i0 < i, j0 < 0, i0 * 256 + j0 < 0, 0 < i0 + i ⊦ i = 0, i2b(256 * (i0 + i) + j0) = i2b(j0), i0 + i = 0

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 256 * (i0 + i) + j0 = (i0 + i - 1) * 256 + 256 + j0

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ i * j + i = i * (j + 1);% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + j = i + i0 ↔ j = i0; Used formulas: 256 * (i0 + i) + j0 = (i0 + i - 1) * 256 + 256 + j0