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Proof step 1 from proof firstN-dups-03 (conclusion)

⊦ n ≤ # y ∧ ¬ dups(x + y) → ¬ dups(x + firstn(n, y))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
dups(x + firstn(n, y)), n ≤ # y ⊦ dups(x + y), dups(x), dups(y)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%a: ⊦ ¬ dups(x + y) → ¬ dups(x) ∧ ¬ dups(y); (forward) from specification list-dup