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Proof step 22 from proof bits2bits-in (previous step)

sgn(lastn(8, bits)) ≠ [], ~128 ≤ bits2int(lastn(8, bits)), bits2int(lastn(8, bits)) ≤ 127, 8 < # bits, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ sgn(lastn(8, bits)) ↔ boolvar ∈ bits, boolvar ∈ firstn(# bits - 8, bits), boolvar ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
~128 ≤ bits2int(lastn(8, bits)), bits2int(lastn(8, bits)) ≤ 127, 8 < # bits, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ sgn(lastn(8, bits)) ↔ boolvar ∈ bits, boolvar ∈ firstn(# bits - 8, bits), boolvar ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits)))

Rule argument

Additional information

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