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Proof step 2 from proof get-firstn (previous step)

n < m, m ≤ # x, n < # x ⊦ firstn(m, x)[n] = x[n], m = 0, x = []

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ n, m. n < m ∧ m ≤ # x0 ∧ n < # x0 → firstn(m, x0)[n] = x0[n] ∨ m = 0 ∨ x0 = [], n < m, m ≤ #(a + x0), n < #(a + x0) ⊦ firstn(m, a + x0)[n] = (a + x0)[n], m = 0, a + x0 = []

Rule argument

term: x

Additional information

Used simplifier rules: list freely generated by [], +;;