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Proof step 108 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 = [], 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

Rule argument

Additional information

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