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Proof step 1 from proof bits2int-neg-03 (conclusion)

⊦ bits.first ∧ bits ≠ [] → bits2int(bits) = ~ bits2uint(bitsinvert(bits) +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bits.first ⊦ bits = [], bits2uint(bitsinvert(bits) +1) + bits2int(bits) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0;