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Proof step 1 from proof bitsand-fillsgn (conclusion)

⊦ bits ≠ [] ∧ bits0 ≠ [] → fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information