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Proof step 1 from proof get-firstn (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < m, m ≤ # x, n < # x ⊦ firstn(m, x)[n] = x[n], m = 0, x = []

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1