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Proof step 1 from proof sublist-prefix (conclusion)

⊦ n ≤ n1 ∧ n1 + m0 < # x → (sublist(m0, n + 1, x) ⊑ sublist(m0, n1 + 1, x) ↔ true)

System applied rule normalize

And obtained the following new goals

Proof step 2:
n1 + m0 < # x, n ≤ n1 ⊦ sublist(m0, n + 1, x) ⊑ sublist(m0, n1 + 1, x) ↔ true

Rule argument

Additional information