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Proof step 1 from proof d (conclusion)

⊦ ¬ a ∈ x ∧ ¬ a ∈ y → (dups(x + a ' + y) ↔ dups(x + y))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ dups(x + a ' + y) ↔ dups(x + y), a ∈ y, a ∈ x

Rule argument

Additional information