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Proof step 1 from proof ubits2ubits-neg (conclusion)

⊦ bits.first ∧ bits ≠ [] → uint2bits(bits2uint(bits)) = false ' + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bits.first ⊦ bits = [], uint2bits(bits2uint(bits)) = false ' + bits

Rule argument

Additional information