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Proof step 4 from proof bits2int-ones-02 (previous step)

¬((bits + bits0).first ∧ bits + bits0 ≠ []) ⊦ bits = [], bits2uint((false ' + bitsinvert(bits0))+1) + bits2int(bits + bits0) = 0, false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ bits = [], bits2uint((false ' + bitsinvert(bits0))+1) + bits2int(bits + bits0) = 0, false ∈ bits, bits.first

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);