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

fillsgn(16, int2bits(s2i(sho))) ≠ [] ∧ fillsgn(16, int2bits(s2i(sho))).first, s2i(sho) < 0 ⊦ bits2int(fillsgn(16, int2bits(s2i(sho)))) + 2 ^ n→i(# fillsgn(16, int2bits(s2i(sho)))) = s2i(sho) + 65536

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;