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Proof step 4 from proof sgn-fill (previous step)

¬(# bits ≤ n ∧ bits ≠ []) ⊦ bits = [], sgn(fillsgn(n, bits)) = sgn(bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < # bits ⊦ sgn(fillsgn(n, bits)) = sgn(bits), bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;