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Proof step 9 from proof bib-ls-same-01 (previous step)

¬ abs(i) < 256, i % 256 + 256 < i, i % 256 < ~128, i < 0 ⊦ i % 256 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i ≤ ~256, i < 0, i % 256 + 256 < i, i % 256 < ~512 ⊦ i % 256 = 0

Rule argument

Additional information

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