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Proof step 4 from proof dups-03 (previous step)

list2set(x) ∩ list2set(y) = ∅, dups(x) ∨ dups(y) ∨ (∃ a. a ∈ x ∧ a ∈ y) ⊦ dups(y), dups(x)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: intempty: ⊦ a ∈ x ∧ a ∈ y → list2set(x) ∩ list2set(y) ≠ ∅;% ⊦ s ∩ s0 = s0 ∩ s;% ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;