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Proof step 7 from proof ubyte-neg (previous step)

fillsgn(8, int2bits(b2i(by))) ≠ [] ∧ fillsgn(8, int2bits(b2i(by))).first, b2i(by) < 0 ⊦ bits2int(fillsgn(8, int2bits(b2i(by)))) + 2 ^ n→i(# fillsgn(8, int2bits(b2i(by)))) = 256 + b2i(by)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ j < 0 → int2bits(j).first;% ⊦ fillsgn(n, int2bits(j)).first ↔ int2bits(j).first;