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Proof step 76 from proof bits2bits-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];%bits2bytes-empty: ⊦ bits2bytes(bits) ≠ [];