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Theorem prefixn-eq-prefixn-app

⊦ prefixn(n0, x0) = prefixn(n1, x0 + x1) ↔ n0 < # x0 ∧ n0 = n1 ∨ n0 ≥ # x0 ∧ n1 ≥ # x0 ∧ x1 = [] ∨ n0 ≥ # x0 ∧ n1 = # x0 ∧ x1 ≠ []

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