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Proof step 11 from proof get-finto (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m0, n0

Additional information

Used terms: m0, n0