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Proof step 22 from proof bits2bits-inj (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
bytes2bits(bits2bytes(firstn(# bits - 8, bits))) = firstn(# bits - 8, bits), # bits % 8 = 0, # lastn(8, bits) = 8, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ firstn(# bits - 8, bits) + lastn(8, bits) = bits, bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ n < m → m - n ≠ 0;% ⊦ firstn(n, bits) = [] ↔ n = 0;