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Proof step 1 from proof fill-lastn-03 (conclusion)

⊦ bits ≠ [] ∧ m ≤ n → lastn(m, fillsgn(n, bits)) = lastn(m, fillsgn(m, bits))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m ≤ n ⊦ bits = [], lastn(m, fillsgn(n, bits)) = lastn(m, fillsgn(m, bits))

Rule argument

Additional information