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Proof step 6 from proof bib-ls-same-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i % 256 + 256 < i, i % 256 < ~128, i < 0 ⊦ i % 256 = 0

Rule argument

Additional information

Used simplifier rules: bib-neg-low: ⊦ i < 0 ∧ i % 256 < ~128 → b2i(i2b(i)) = i % 256 + 256;% ⊦ i < 0 ∧ 0 ≤ j → i < j;%0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;