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Proof step 1 from proof fill-lastn (conclusion)

⊦ n < # bits → lastn(n, fillsgn(n, bits)) = lastn(n, bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < # bits ⊦ lastn(n, fillsgn(n, bits)) = lastn(n, bits), bits = []

Rule argument

Additional information

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