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Proof step 6 from proof append-to-le (previous step)

n0 ≤ # x0, ∀ x0, a, y. n0 ≤ # x0 + 1 ∧ (∀ n, y. n ≤ # x0 → x0 + y to n = x0 to n) → (a ' + x0) + y to n0 = a ' + x0 to n0, ∀ n, y. n ≤ # x0 → x0 + y to n = x0 to n ⊦ x0 + y to n0 = x0 to n0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n0, y

Additional information

Used terms: n0, y