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Proof step 24 from proof bib-ls-same-01 (previous step)

¬ abs(i) < 256, 0 < i % 256, 0 ≤ i, 127 < i % 256 ⊦ i % 256 < 256 + i ↔ ¬ i < 128

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
0 < i % 256, 0 ≤ i, 127 < i % 256, 256 ≤ i, 256 + i ≤ i % 256, 0 < i ⊦

Rule argument

Additional information

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