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Proof step 4 from proof ubits2bits (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
bits.first ⊦ bits = [], 2 ^ n→i(# bits) = bits2uint(bits) + bits2uint(bitsinvert(bits) +1)

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;