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Proof step 1 from proof bits2int-ones-02 (conclusion)

⊦ ¬ false ∈ bits ∧ bits ≠ [] → bits2int(bits + bits0) = bits2int(true ' + bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ bits = [], bits2uint((false ' + bitsinvert(bits0))+1) + bits2int(bits + bits0) = 0, false ∈ bits

Rule argument

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits2int(true ' + bits) = ~ bits2uint(bitsinvert(true + bits) +1);% ⊦ ~ i = j ↔ i + j = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ boolvar + bits = boolvar ' + bits;