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Proof step 11 from proof bits2bits-sgn (previous step)

~128 ≤ bits2int(bits) ∧ bits2int(bits) ≤ 127, # bits ≤ 8, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [], sgn(fillsgn(8, int2bits(bits2int(bits)))) = sgn(bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
~128 ≤ bits2int(bits), bits2int(bits) ≤ 127, # bits ≤ 8, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [], sgn(mklist(bits.first, 8 - # bits) + bits) = sgn(bits)

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)%⊦ # bits ≤ n ∧ bits ≠ [] → fillsgn(n, bits) = mklist(bits.first, n - # bits) + bits;% ⊦ # bits ≤ n → fillsgn(n, sgn(bits)) = fillsgn(n, bits);%bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);