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Proof step 24 from proof bits2bits-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ n < m → m - n ≠ 0;% ⊦ firstn(n, bits) = [] ↔ n = 0;