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Proof step 7 from proof dups-01 (previous step)

⊦ dups(x), ¬(dups(x) ∨ dups(set2list(s)) ∨ (∃ a. a ∈ x ∧ a ∈ set2list(s))) ∧ ¬ dups(x + set2list(s0)), ∃ a. a ∈ x ∧ (a ∈ s ∨ a ∈ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ dups(x), ∃ a. a ∈ x ∧ (a ∈ s ∨ a ∈ s0), (¬ a ∈ s ∨ ¬ a ∈ x) ∧ ¬ dups(x + set2list(s0))

Rule argument

Additional information

Used simplifier rules: dups: ⊦ ¬ dups(set2list(s));%in-01: ⊦ a ∈ set2list(s) ↔ a ∈ s;