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Proof step 1 from proof uint2uint-neg (conclusion)

⊦ i < 0 → bits2uint(uint2bits(~ i)) = ~ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i < 0 ⊦ i + bits2uint(uint2bits(~ i)) = 0

Rule argument

Additional information

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