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

n + m0 ≤ # x0, ∀ n, x0, a. n + m0 ≤ # x0 + 1 ∧ (∀ n, m. n + m ≤ # x0 → x0 from m from n = x0 from m + n) → a ' + x0 from m0 from n = a ' + x0 from m0 + n, ∀ n, m. n + m ≤ # x0 → x0 from m from n = x0 from m + n ⊦ x0 from m0 from n = x0 from m0 + n

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n, m0

Additional information

Used terms: n, m0