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Proof step 44 from proof bits2bits-sgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
firstn(# bits - 8, bits) + lastn(8, bits) = bits, 8 < # bits, 0 ≤ b2i(i2b(bits2int(lastn(8, bits)))), ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [], sgn(false ' + byte2bits(i2b(bits2int(lastn(8, bits))))) = sgn(false ' + lastn(8, bits)), true ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits))), true ∈ firstn(# bits - 8, bits)

Rule argument

Additional information

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