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Proof step 22 from proof bits2bytes-fill (previous step)

# bits % 8 ≤ 8 ∧ 8 ≤ 8, ((8 - # bits % 8) + # bits) % 8 = 0, 8 < (8 - # bits % 8) + # bits ⊦ 8 + # bits = # bits % 8, bits = [], # bits % 8 = 0, bits2bytes(bits) = bits2bytes(firstn(8, mklist(bits.first, 8 - # bits % 8) + bits)) + bits2bytes(restn((8 - 8) + # bits % 8, bits))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
((8 - # bits % 8) + # bits) % 8 = 0, 8 < (8 - # bits % 8) + # bits ⊦ 8 + # bits = # bits % 8, bits = [], # bits % 8 = 0, bits2bytes(bits) = bits2bytes(firstn(8, mklist(bits.first, 8 - # bits % 8) + bits)) + bits2bytes(restn(# bits % 8, bits))

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m % n ≤ n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;