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Proof step 6 from proof get-firstn (previous step)

n < m, m ≤ 1 + # x0, n ≤ # x0, ∀ n, m. n < m ∧ n < # x0 ∧ m ≤ # x0 → firstn(m, x0)[n] = x0[n] ∨ x0 = [] ⊦ firstn(m, a ' + x0)[n] = (a ' + x0)[n], x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n < m, n ≤ # x0, m ≤ 1 + # x0, ∀ n0, m0. n0 < m0 ∧ n0 < # x0 ∧ m0 ≤ # x0 → firstn(m0, x0)[n0] = x0[n0] ⊦ firstn(m, a ' + x0)[n] = (a ' + x0)[n], x0 = []

Proof step 6 from proof get-firstn (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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