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Proof step 6 from proof to-plus-from-fin (previous step)

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