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Proof step 6 from proof bits2bytes-rec-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
8 < # bits ⊦ # bits % 8 = 0, bits2bytes(firstn(# bits % 8, bits) + restn(# bits % 8, bits)) = bits2bytes(firstn(# bits % 8, bits)) + bits2bytes(restn(# bits % 8, bits)), bits = [], (# bits - # bits % 8) % 8 = 0 ∧ restn(# bits % 8, bits) ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ firstn(n, bits) = [] ↔ n = 0;%n ≤ # bits ⊦ # restn(n, bits) = # bits - n;%n ≠ 0 ⊦ m % n ≤ m;