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Proof step 14 from proof bib-ls-same (previous step)

¬(0 ≤ i ∧ 127 < i % 256), i < b2i(i2b(i)), 0 ≤ i, 0 < b2i(i2b(i)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i % 256 ≤ 127, i < i % 256, 0 ≤ i, i < 127, 0 < i % 256 ⊦

Rule argument

Additional information

Used simplifier rules: bib-pos-low: ⊦ 0 ≤ i ∧ i % 256 ≤ 127 → b2i(i2b(i)) = i % 256;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;