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Proof step 65 from proof bits2bits-sgn (previous step)

lastn(8, bits) ≠ [] ∧ (lastn(8, bits).first ↔ ¬ false), firstn(# bits - 8, bits) + lastn(8, bits) = bits, 8 < # bits, b2i(i2b(bits2int(lastn(8, bits)))) < 0, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [], false ' + byte2bits(i2b(bits2int(lastn(8, bits)))) = 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 66:
firstn(# bits - 8, bits) + lastn(8, bits) = bits, 8 < # bits, b2i(i2b(bits2int(lastn(8, bits)))) < 0, lastn(8, bits).first, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [], byte2bits(i2b(bits2int(lastn(8, bits)))) = lastn(8, bits), true ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits))), true ∈ firstn(# bits - 8, bits)

Rule argument

Additional information

Used simplifier rules: ⊦ lastn(n, bits) = [] ↔ n = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;