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Proof step 19 from proof bits2bits-inj (previous step)

8 ≠ 0, ∃ n0. # bits = n0 * 8, # 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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# lastn(8, bits) = 8, # bits = n * 8, 8 < n * 8, ∀ bits0. # bits0 < n * 8 → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ bits = [], bytes2bits(bits2bytes(firstn(n * 8 - 8, bits))) + lastn(8, bits) = bits, (n * 8 - 8) % 8 = 0, n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)