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Proof step 77 from proof byte-mult-256 (previous step)

(i0 + i + 1) * 256 + j0 = i1 + 256, i0 + i = ~1, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i = 0, i2b(i1) = i2b(j0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 78:
i0 + i = ~1, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i2b(j0 - 256) = i2b(j0), i = 0

Rule argument

Additional information

Used simplifier rules: i2b-noteq: ⊦ i2b(i) ≠ i2b(j) → i ≠ j;% ⊦ i = i + j ↔ j = 0;% ⊦ i - j = i0 ↔ i = i0 + j;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i * (i0 + 1) + j = i + j0 ↔ i * i0 + j = j0;