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Proof step 3 from proof bits2int-split (previous step)

¬ (bits + bits0).first ∧ bits + bits0 ≠ [] ⊦ # bits0 ≠ 0 ∧ # bits ≠ 0 ∧ ¬ bits.first → bits2uint(bits + bits0) = bits2uint(bits) * 2 ^ n→i(# bits0) + bits2uint(bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bits + bits0 = [], bits2uint(bits + bits0) = bits2uint(bits) * 2 ^ n→i(# bits0) + bits2uint(bits0), bits = [], bits0 = [], bits.first

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ # bits = 0 ↔ bits = [];