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Proof step 7 from proof ubits2bits (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
bits.first ⊦ bits = [], 2 ^ n→i(# bits) = bits2uint(bits) + bits2uint(bitsinvert(bits)) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ bitsinvert(bits) = [] ↔ bits = [];