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Proof step 82 from proof bits2bits-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 83:
8 < # bits, b2i(i2b(bits2int(lastn(8, bits)))) < 0, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [], sgn(true ' + byte2bits(i2b(bits2int(lastn(8, bits))))) = sgn(bits), false ∈ firstn(# bits - 8, bits), false ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits)))

Rule argument

Additional information

Used simplifier rules: byte2bits-empty: ⊦ byte2bits(by) ≠ [];%byte2bits-first: ⊦ byte2bits(by).first ↔ b2i(by) < 0;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)