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Proof step 3 from proof d (previous step)

⊦ dups(x + a ' + y) ↔ dups(x + y), a ∈ x, a ∈ y

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ a, y. (dups(x0 + a ' + y) ↔ dups(x0 + y)) ∨ a ∈ x0 ∨ a ∈ y ⊦ dups((a0 + x0) + a ' + y) ↔ dups((a0 + x0) + y), a ∈ y, a ∈ a0 + x0

Rule argument

term: x

Additional information

Used simplifier rules: dups-cons: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x;%list freely generated by [], +;;% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list