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Proof step 12 from proof get-firstn (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m, m0

Additional information

Used terms: m, m0