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Proof step 4 from proof bib-pos-high (previous step)

¬(~128 ≤ i ∧ i ≤ 127), 0 < i % 256, 127 < i % 256, 0 ≤ i ⊦ i % 256 = b2i(i2b(i)) + 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i % 256, 127 < i % 256, 127 < i, 0 < i, 0 ≤ i ⊦ i % 256 = b2i(i2b(i)) + 256

Rule argument

Additional information

Used simplifier rules: ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ ¬ i ≤ j ↔ j < i;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;