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Proof step 8 from proof to-from-01 (previous step)

n0 < # x0, ∀ x0, a. n0 ≤ # x0 ∧ (∀ n. n < # x0 → (x0 to n) from n = [] ∨ x0 = []) → (a ' + x0 to n0) from n0 = [], ∀ n. n < # x0 → (x0 to n) from n = [] ⊦ (x0 to n0) from n0 = [], x0 = []

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n0

Additional information

Used terms: n0