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Proof step 6 from proof bits2int-split-01 (previous step)

¬ ¬ true ∈ bits0 ⊦ bits0 = [], bits2uint((false ' + bitsinvert(bits1) + 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 7:
true ∈ bits0 ⊦ bits0 = [], bits2uint(bitsinvert(bits1) + bitsinvert(bits0) +1) = (bits2uint(bitsinvert(bits1)) * 2 ^ n→i(# bits0) + bits2uint(bitsinvert(bits0))) + 1

Rule argument

Additional information

Used simplifier rules: bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ true ∈ bits0 → (bits + bitsinvert(bits0))+1 = bits + bitsinvert(bits0) +1;