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Proof step 1 from proof bits2bytes-rec-01 (conclusion)

⊦ # bits % 8 ≠ 0 ∧ 8 < # bits → bits2bytes(bits) = bits2bytes(firstn(# bits % 8, bits)) + bits2bytes(restn(# bits % 8, bits))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
8 < # bits ⊦ # bits % 8 = 0, bits2bytes(bits) = bits2bytes(firstn(# bits % 8, bits)) + bits2bytes(restn(# bits % 8, bits)), bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)