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Proof step 6 from proof bits2bytes-fill (previous step)

# bits = m * 8 + n ∧ n < 8 ∧ 8 ≠ 0 ⊦ bits2bytes(bits) = bits2bytes(mklist(bits.first, 8 - n) + bits), n = 0, bits = [], ((8 - n) + # bits) % 8 = 0 ∧ ¬ # bits ≤ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bits = m * 8 + n, n < 8 ⊦ bits = [], n = 0, bits2bytes(bits) = bits2bytes(mklist(bits.first, 8 - n) + bits), ((8 - n) + # bits) % 8 = 0 ∧ ¬ # bits ≤ n

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)