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Proof step 12 from proof bits2int-split-01 (previous step)

bitsinvert(bits0) ≠ [], 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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
true ∈ bits0 ⊦ bitsinvert(bits0) +1 = [], bits0 = [], bits2uint(bitsinvert(bits1)) * 2 ^ n→i(# bitsinvert(bits0) +1) = bits2uint(bitsinvert(bits1)) * 2 ^ n→i(# bits0)

Rule argument

Additional information

Used simplifier rules: ⊦ bitsinvert(bits) = [] ↔ bits = [];% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = i + i0 ↔ j = i0;