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Proof step 12 from proof bib-pos-high (previous step)

0 < i, 0 < i % 256, 127 < i % 256, 127 < i, 0 < i ⊦ i % 256 = b2i(i2b(bits2int(lastn(8, int2bits(i))))) + 256, 0 = i, lastn(8, int2bits(i)).first ∧ ¬ i < 2 ^ (n→i(8) - 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ ¬ i < j ↔ j ≤ i;