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Proof step 9 from proof bits2int-split-01 (previous step)

# bitsinvert(bits0) +1 ≠ 0, true ∈ bits0 ⊦ bits0 = [], bits2uint(bitsinvert(bits1)) * 2 ^ n→i(# bitsinvert(bits0) +1) + bits2uint(bitsinvert(bits0) +1) = (bits2uint(bitsinvert(bits1)) * 2 ^ n→i(# bits0) + bits2uint(bitsinvert(bits0))) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
true ∈ bits0 ⊦ bits0 = [], bits2uint(bitsinvert(bits1)) * 2 ^ n→i(# bitsinvert(bits0) +1) + bits2uint(bitsinvert(bits0) +1) = (bits2uint(bitsinvert(bits1)) * 2 ^ n→i(# bits0) + bits2uint(bitsinvert(bits0))) + 1, bitsinvert(bits0) +1 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];