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Proof step 14 from proof bits2bytes-fill (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
m * 8 = 0, # bits < 8 ⊦ bits = [], i2b(bits2int(bits)) = i2b(bits2int(mklist(bits.first, 8 - # bits) + bits)), bits2int(bits) = bits2int(mklist(bits.first, 8 - # bits) + bits)

Rule argument

Additional information

Used simplifier rules: ⊦ i2b(i) ≠ i2b(j) → i ≠ j; (forward)% ⊦ m < n → m ≤ n;% ⊦ by ' = by0 ' ↔ by = by0;