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

¬ n ≤ # bits, fillsgn(n, bits) = [] ⊦ bits = [] ∧ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n → n ≠ 0;% ⊦ ¬ m ≤ n ↔ n < m;