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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < m, n < # x, m ≤ # x ⊦ firstn(m, x)[n] = x[n], x = []

Rule argument

Additional information

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