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Proof step 2 from proof d (previous step)

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ a, y. (dups(x0 + a ' + y) ↔ dups(x0 + y)) ∨ a ∈ y ∨ a ∈ x0 ⊦ dups((a0 + x0) + a ' + y) ↔ dups((a0 + x0) + y), a ∈ y, a ∈ a0 + x0

Rule argument

term: x

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%app-01-02: ⊦ dups(y) → dups(x + y);%list freely generated by [], +;;%Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list