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Proof step 97 from proof bits2bits-sgn (previous step)

# lastn(8, bits) = 8, ~128 ≤ bits2int(lastn(8, bits)), 8 < # bits, bits2int(lastn(8, bits)) < 0, b2i(i2b(bits2int(lastn(8, bits)))) < 0, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [], sgn(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 98:
# lastn(8, bits) = 8, b2i(i2b(bits2int(lastn(8, bits)))) < 0, bits2int(lastn(8, bits)) < 0, 8 < # bits, ~128 ≤ bits2int(lastn(8, bits)), ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ sgn(lastn(8, bits)) = sgn(bits), bits = [], false ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits))), false ∈ firstn(# bits - 8, bits)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ m = n0 → n0 < n; (forward)