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Proof step 18 from proof byte-mult-256 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 19:
256 * i + j0 = (i - 1) * 256 + 256 + j0, j0 < 0, ~256 < j0, 0 < i ⊦ i2b(256 * i + j0) = i2b(j0), i = 0

Rule argument

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

Additional information

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