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Proof step 1 from proof dups-firstoc-rew (conclusion)

⊦ dups(x) ↔ (∃ y, a, z. x = y + a ' + z ∧ a ∈ y ∧ ¬ dups(y))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
dups(x) ⊦ ∃ y, a, z. x = y + a ' + z ∧ a ∈ y ∧ ¬ dups(y)

Rule argument

Additional information

Used simplifier rules: in-04-01: ⊦ a ∈ x → dups(x + a ' + y);%a: ⊦ (x + y) + z = x + y + z; from specification list