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Proof step 58 from proof byte-mult-256 (previous step)

j < 0, 0 < i ⊦ j % 256 < 0, i = 0, i2b(i * 256 + j) = i2b(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 59:
j % 256 = 0, j < 0, 0 < i ⊦ i = 0, i2b(i * 256 + j) = i2b(j)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ 0 < j ∧ i < 0 → (0 ≤ i % j ↔ i % j = 0);% ⊦ ¬ i < j ↔ j ≤ i;