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Proof step 2 from proof in-06 (previous step)

a ∈ x, a ∈ y ⊦ dups(x + y)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a, y. a ∈ x0 ∧ a ∈ y → dups(x0 + y), a ∈ a0 + x0, a ∈ y ⊦ dups((a0 + x0) + y)

Rule argument

term: x

Additional information

Used simplifier rules: list freely generated by [], +;;%in-empty: ⊦ ¬ a ∈ []; from specification list