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Proof step 16 from proof bits2bits-inj (previous step)

# lastn(8, bits) = 8, # bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ # firstn(# bits - 8, bits) % 8 = 0, bits = [], bytes2bits(bits2bytes(firstn(# bits - 8, bits))) + lastn(8, bits) = bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# bits % 8 = 0, # lastn(8, bits) = 8, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ (# bits - 8) % 8 = 0, bytes2bits(bits2bytes(firstn(# bits - 8, bits))) + lastn(8, bits) = bits, bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ # firstn(n, bits) = n;