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Proof step 4 from proof bib-lower (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
~256 < i, i < ~128, i < 0, ~128 ≤ i % 256, i < i % 256 ⊦ i % 256 = i + 256

Rule argument

Additional information

Used simplifier rules: bib-neg-high: ⊦ i < 0 ∧ ~128 ≤ i % 256 → b2i(i2b(i)) = i % 256;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;