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Proof step 6 from proof firstN-dups-04 (
previous step
)
n0 ≤ # y, dups(firstn(n0, y)), ∀ x. dups(firstn(n0, x)) ∧ n0 ≤ # x → dups(x) ⊦ a ∈ y, dups(y)
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: y
Additional information
Used terms: y