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

¬ n ≤ # bits ⊦ bits = [], boolvar ∈ fillsgn(n, bits) ↔ boolvar ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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