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Proof step 3 from proof get-firstn (previous step)

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
n < #(a + x0), n < m, m ≤ #(a + x0), ∀ n, m. n < m ∧ n < # x0 ∧ m ≤ # x0 → firstn(m, x0)[n] = x0[n] ∨ x0 = [] ⊦ a + x0 = [], firstn(m, a + x0)[n] = (a + x0)[n]

Rule argument

term: x

Additional information

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