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Proof step 8 from proof byte-law-add-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
~256 < j0, i0 ≤ 0, j0 < 0, i0 * 256 + j0 < 0 ⊦ i2b(j0 + j) = i2b(b2i(i2b(j0)) + j)

Rule argument

Additional information

Used simplifier rules: byte-mult-256: ⊦ i2b(i * 256 + j) = i2b(j);% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;