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Proof step 4 from proof dups-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ dups(x), ¬ dups(x + set2list(s)) ∧ ¬ dups(x + set2list(s0)), ∃ a. a ∈ x ∧ (a ∈ s ∨ a ∈ s0)

Rule argument

Additional information

Used simplifier rules: dups: ⊦ ¬ dups(set2list(s));%in-01: ⊦ a ∈ set2list(s) ↔ a ∈ s;% ⊦ s ∪ s0 = s0 ∪ s;% ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;% ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2;