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Proof step 5 from proof get-firstn (previous step)

n ≤ # x0, m ≤ # x0 + 1, n < m, ∀ n, m. n < m ∧ m ≤ # x0 ∧ n < # x0 → firstn(m, x0)[n] = x0[n] ∨ m = 0 ∨ x0 = [] ⊦ x0 = [], firstn(m, a ' + x0)[n] = (a ' + x0)[n], m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n ≤ # x0, m ≤ # x0 + 1, n < m, ∀ n0, m0. n0 < m0 ∧ m0 ≤ # x0 ∧ n0 < # x0 → firstn(m0, x0)[n0] = x0[n0] ∨ m0 = 0 ⊦ x0 = [], firstn(m, a ' + x0)[n] = (a ' + x0)[n], m = 0

Proof step 5 from proof get-firstn (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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