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Proof step 69 from proof fix-bits2bytes-01 (previous step)

n0 ≠ # bits2bytes(bits), # bits = 8 * n0, n0 ≤ # bits2bytes(bits) ⊦ n0 = 0, bits = [], lastn(n0, bits2bytes(bits)) = bits2bytes(bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
# bits = 8 * n0, n0 < # bits2bytes(bits) ⊦ n0 = # bits2bytes(bits), lastn(n0, bits2bytes(bits)) = bits2bytes(bits), bits = [], n0 = 0

Rule argument

Additional information

Used simplifier rules: bits2bytes-empty: ⊦ bits2bytes(bits) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m ≠ n → (m ≤ n ↔ m < n);