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Proof step 4 from proof bits2uint-zeros-02 (previous step)

¬ # bits0 ≠ 0 ⊦ bits2uint(bits + bits0) = bits2uint(bits0), true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ bits2uint(bits) = 0, true ∈ bits

Rule argument

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ # bits = 0 ↔ bits = [];