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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;% ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ i - j = i0 ↔ i = i0 + j;