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Proof step 6 from proof bits2bits-inj (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ bits = [], bytes2bits(bits2bytes(bits)) = bits

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;