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Proof step 85 from proof bits2bits-sgn (previous step)

~128 ≤ bits2int(lastn(8, bits)) ∧ bits2int(lastn(8, bits)) ≤ 127, 8 < # bits, 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)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;