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Proof step 13 from proof byte-mult-neg (previous step)

i * 256 + j < 0 ∧ (i * 256 + j) % 256 < ~128, ¬(i * 256 + j < 0 ∧ ~128 ≤ (i * 256 + j) % 256), i < 0, j < 0 ⊦ (i * 256 + j) % 256 + 256 = b2i(i2b(j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i * 256 + j < 0, j < 0, i < 0, (i * 256 + j) % 256 < ~128 ⊦ (i * 256 + j) % 256 + 256 = b2i(i2b(j)), (i * 256 + j) % 256 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;%0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;