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Proof step 35 from proof bits2bytes-app (previous step)

# bits0 % 8 = 0, 8 < # bits0, 8 < # bits + # bits0, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ # firstn(# bits0 - 8, bits0) % 8 = 0, bits0 = [], bits = [], bits2bytes(bits + firstn(# bits0 - 8, bits0)) = bits2bytes(bits) + bits2bytes(firstn(# bits0 - 8, bits0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
# bits0 % 8 = 0, 8 < # bits0, 8 < # bits + # bits0, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ bits0 = [], bits = [], bits2bytes(bits + firstn(# bits0 - 8, bits0)) = bits2bytes(bits) + bits2bytes(firstn(# bits0 - 8, bits0)), (# bits0 - 8) % 8 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ # firstn(n, bits) = n;