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Proof step 1 from proof bib-pos-low (conclusion)

⊦ 0 ≤ i ∧ i % 256 ≤ 127 → b2i(i2b(i)) = i % 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i % 256 ≤ 127, 0 ≤ i ⊦ b2i(i2b(i)) = i % 256

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)