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Proof step 1 from proof sublist-at (conclusion)

⊦ n0 < n ∧ m + n ≤ # x → sublist(m, n, x)[n0] = x[m + n0]

System applied rule normalize

And obtained the following new goals

Proof step 2:
n0 < n, m + n ≤ # x ⊦ sublist(m, n, x)[n0] = x[m + n0]

Rule argument

Additional information