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Proof step 56 from proof bits2bits-sgn (previous step)

# lastn(8, bits) ≠ 8, firstn(# bits - 8, bits) + lastn(8, bits) = bits, bits2int(lastn(8, bits)) ≤ 127, 8 < # bits, 0 ≤ bits2int(lastn(8, bits)), ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [], sgn(byte2bits(i2b(bits2int(lastn(8, bits))))) = sgn(lastn(8, bits)), true ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits))), true ∈ firstn(# bits - 8, bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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