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Proof step 1 from proof isprefix-prefixn (conclusion)

⊦ x0 ⊑ x1 → (x0 = prefixn(n, x1) ↔ n = # x0 ∨ n > # x0 ∧ x0 = x1)

System applied rule normalize

And obtained the following new goals

Proof step 2:
x0 ⊑ x1 ⊦ x0 = prefixn(n, x1) ↔ n = # x0 ∨ n > # x0 ∧ x0 = x1

Rule argument

Additional information