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Proof step 4 from proof bits2uint-inv-neg (previous step)

¬ bits.first ∧ bits ≠ [] ⊦ bits = [], 2 ^ n→i(# bits) = bits2uint(bitsinvert(bits)) + bits2uint(bits) + 1, bits.first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information