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Proof step 38 from proof bits2bytes-fill (previous step)

# firstn(# bits % 8, bits) ≤ 8, ((8 - # bits % 8) + # bits) % 8 = 0, 8 < (8 - # bits % 8) + # bits, 8 < # bits ⊦ i2b(bits2int(firstn(# bits % 8, bits))) ' = bits2bytes(firstn(8, mklist(bits.first, 8 - # bits % 8) + bits)), bits = [], 8 + # bits = # bits % 8, # bits % 8 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
((8 - # bits % 8) + # bits) % 8 = 0, 8 < (8 - # bits % 8) + # bits, 8 < # bits ⊦ i2b(bits2int(firstn(# bits % 8, bits))) ' = bits2bytes(firstn(8, mklist(bits.first, 8 - # bits % 8) + bits)), bits = [], 8 + # bits = # bits % 8, # bits % 8 = 0

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m % n ≤ n;% ⊦ # firstn(n, bits) = n;