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Proof step 6 from proof firstn-restn-eq (previous step)

n0 ≤ # y, ∀ x. n0 ≤ # x → firstn(n0, x) + restn(n0, x) = x ⊦ firstn(n0, y) + restn(n0, y) = y

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: y

Additional information

Used terms: y