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Proof step 19 from proof bits2bits-in (previous step)

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 ∈ fillsgn(8, 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 20:
~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 ∈ fillsgn(8, 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: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ lastn(n, bits) = [] ↔ n = 0;