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Proof step 25 from proof bits2bits-inj (previous step)

# bits ≤ 8, # bits % 8 = 0, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ bytes2bits(i2b(bits2int(bits)) ') = bits, bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# bits % 8 = 0, # bits ≤ 8, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ bits = [], byte2bits(i2b(bits2int(bits))) = bits

Rule argument

Additional information

Used simplifier rules: bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);