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Proof step 15 from proof bint-and-011x-bits2unit (previous step)

fillsgn(32, int2bits(i)) ≠ [] ∧ fillsgn(32, int2bits(i)).first, ~2147483648 ≤ i, i < 0 ⊦ bits2int(fillsgn(32, int2bits(i))) + 2 ^ n→i(# fillsgn(32, int2bits(i))) = 4294967296 + i

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;