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Proof step 48 from proof bib-neg-high (previous step)

i0 < 0 ⊦ i0 * 256 - 128 = (2 * i0 - 1) * 2 ^ 7, b2i(i2b(bits2int(lastn(8, int2bits(i0 * 256 - 128))))) = ~128, i0 * 256 ≠ ~128 ∧ lastn(8, int2bits(i0 * 256 - 128)).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs