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Proof step 1 from proof f (conclusion)

⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ a. dups(x0 + a ') ↔ a ∈ x0 ∨ dups(x0) ⊦ dups((a0 + x0) + a ') ↔ a ∈ a0 + x0 ∨ dups(a0 + x0)

Rule argument

term: x

Additional information

Used simplifier rules: c: ⊦ ¬ dups(a ');%e-01: ⊦ ¬ dups([]);%list freely generated by [], +;;%Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-empty: ⊦ ¬ a ∈ []; from specification list