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Proof step 13 from proof get-firstn (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: m, m0

Additional information

Used terms: m, m0